#1 of 40
SZABMU 2024
Permalink
In which of the following condition, the thermodynamics system does not perform any work?
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
1. Concept:In physics, thermodynamic work requires a spatial change. Without a change in boundary (volume), no physical work can be transferred to the environment.
2. Formula:$$W = P \Delta V$$
3. Solution:- To perform zero work, the formula requires \(\Delta V = 0\).
- A process where the volume is artificially or physically locked (such as heating a gas inside a thick, unyielding metal sphere) guarantees \(\Delta V = 0\).
- This rigidly constrained process is officially termed an Isochoric condition.
4. Why other options are incorrect:All other listed processes (isothermal, isobaric, adiabatic) inherently involve a changing volume, meaning they mathematically will produce non-zero work.
#2 of 40
KMU 2024
Permalink
Consider an ideal gas confined to the cylinder with a fixed position, on heating the gas, all the heat supplied increases
A
Kinetic energy of the molecules
B
The intermolecular forces between gas molecules
C
Potential energy of the molecules
D
The number of gas molecules
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
1. Concept:A 'fixed position' piston creates an Isochoric (constant volume) process. Furthermore, an 'ideal gas' has no intermolecular forces, meaning it possesses zero potential energy.
2. Formula:$$Q = \Delta U \quad \text{(where } U = K.E_{\text{translational}} \text{)}$$
3. Solution:- Because the volume is fixed, work done is zero (\(W = 0\)). Thus, First Law states \(Q = \Delta U\).
- All supplied heat goes directly into increasing the internal energy.
- By the kinetic theory of gases, the internal energy of a purely ideal gas consists exclusively of the translational kinetic energy of its molecules.
4. Why other options are incorrect:An ideal gas is theoretically defined as having absolutely no intermolecular forces, therefore zero potential energy. Heating a gas cannot create new molecules.
#3 of 40
UHS 2024
Permalink
What is the value of heat (Q) in an adiabatic process?
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
1. Concept:The cornerstone definition of an adiabatic thermodynamic process is absolute thermal isolation from the surrounding environment.
2. Formula:$$Q = 0$$
3. Solution:- In an adiabatic process, the boundary of the system is considered perfectly insulated (like a perfect thermos flask).
- Because no heat can permeate this boundary, the net heat transfer entering or leaving the system is strictly zero.
- Consequently, \(Q = 0\).
4. Why other options are incorrect:Values like +1, -1, or +2 imply a finite, quantifiable heat transfer, which directly contradicts the physical definition of 'adiabatic'.
#4 of 40
NUMS 2024
Permalink
A process in which all the heat energy is used for increasing internal energy of the system is known as:
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
1. Concept:If all incoming heat strictly increases internal energy, it means none of the energy is being spent performing mechanical work on the environment.
2. Formula:$$Q = \Delta U + W \implies Q = \Delta U \quad \text{if } W=0$$
3. Solution:- For work to be absolute zero, the system's boundary must not move (i.e., volume remains perfectly constant).
- A process occurring strictly at constant volume is formally termed an Isochoric process.
- Therefore, in an isochoric heating process, \(100\%\) of the input heat converts into internal kinetic energy.
4. Why other options are incorrect:In isobaric, part of the heat does work. In isothermal, all heat does work. In adiabatic, no heat enters at all.
#5 of 40
UHS 2023
Permalink
For an ideal gas equation \(PV = nRT\), the dimensions of real gas constant R are:
A
\([M L^{-2} T^{-1} K^{-1}]\)
B
\([M L^2 T^{-2} K^{-1}]\)
C
\([M L^{-3} T^{-2} K^{-1}]\)
D
\([M L^{-3} T^{-1} K^{-1}]\)
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
1. Concept:Dimensional analysis is used to find the base SI units of a constant by isolating it in a known physical equation.
2. Formula:$$R = \frac{PV}{nT}$$
3. Solution:- The product of Pressure and Volume (\(PV\)) has the exact same dimensions as Energy or Work (Joules).
- The dimension of Energy is \([M L^2 T^{-2}]\).
- Temperature (\(T\)) has the dimension of \([K]\). Moles (\(n\)) are dimensionless or defined as \([mol]\), but often omitted in basic dimensional brackets.
- Dividing Energy by Temperature yields: \([M L^2 T^{-2} K^{-1}]\).
4. Why other options are incorrect:Other dimensional formulas represent incorrect physics, such as dividing by volume instead of multiplying (yielding negative length dimensions).
#6 of 40
UHS 2023
Permalink
An ideal gas has molar specific heat \(C_p\) at constant pressure. When the temperature of n moles is increased by \(\Delta T\) the increase in the internal energy is:
B
\(n (C_p - R) \Delta T\)
C
\(n (C_p + R) \Delta T\)
D
\(n (2C_p + R) \Delta T\)
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
1. Concept:The change in internal energy of an ideal gas strictly relies on its molar specific heat at constant volume (\(C_v\)), regardless of the thermodynamic process it undergoes.
2. Formula:$$\Delta U = n C_v \Delta T \quad \text{and} \quad C_p - C_v = R$$
3. Solution:- We know internal energy depends on \(C_v\), so \(\Delta U = n C_v \Delta T\).
- Using Mayer's relation, we can express \(C_v\) in terms of \(C_p\): \(C_v = C_p - R\).
- Substituting this back into the internal energy equation gives: \(\Delta U = n (C_p - R) \Delta T\).
4. Why other options are incorrect:Option \(n C_p \Delta T\) represents the total heat supplied (\(Q\)) at constant pressure, not just internal energy. Other options use incorrect mathematical variations of Mayer's relation.
#7 of 40
SZABMU 2023
Permalink
The relation between \(C_p\) and \(C_v\) is given as:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
1. Concept:Mayer's relation dictates that a gas heated at constant pressure requires more heat to raise its temperature than at constant volume because extra energy is needed to do expansion work against atmospheric pressure.
2. Formula:$$C_p = C_v + R$$
3. Solution:- \(C_p\) (heat capacity at constant pressure) must account for both internal energy and work done.
- \(C_v\) (heat capacity at constant volume) only accounts for internal energy.
- The difference is exactly the work done per mole per Kelvin, which is the universal gas constant \(R\).
- Thus, algebraically: \(C_p - C_v = R\).
4. Why other options are incorrect:\(C_v - C_p = R\) would falsely imply that \(C_v\) is larger. Adding them together has no physical thermodynamic significance.
#8 of 40
SZABMU 2023
Permalink
The equation of \(1^{st}\) law of thermodynamics is:
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
1. Concept:The First Law of Thermodynamics serves as the energy conservation ledger for thermal systems, tracking heat input, energy storage, and work output.
2. Formula:$$Q = \Delta U + W$$
3. Solution:- By standard convention (where work done BY the gas is positive), the heat energy supplied to the system (\(Q\)) serves two simultaneous purposes:
- First, it increases the kinetic energy of the molecules, thus raising the internal energy (\(\Delta U\)).
- Second, any remaining energy pushes the piston outward, performing mechanical work (\(W\)) on the environment.
- Therefore, the total input equals the sum of the outcomes: \(Q = \Delta U + W\).
4. Why other options are incorrect:Equations like \(W = Q + \Delta U\) would violate energy conservation by suggesting work output is greater than heat input when internal energy also increases.
#9 of 40
SZABMU 2023
Permalink
The thermodynamics process during which the pressure is kept constant is called:
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
1. Concept:A thermodynamic process that involves changing state variables (V, T) while explicitly maintaining a constant pressure constraint is known as an isobaric process.
2. Formula:$$P = \text{constant} \implies \Delta P = 0$$
3. Solution:- 'Iso' stands for equal/constant. 'Baric' refers to pressure (originating from 'baros' meaning weight, related to barometers).
- When a gas expands but the external pressure (like a freely moving piston under constant atmospheric weight) doesn't change, the process is Isobaric.
4. Why other options are incorrect:Isochoric means constant volume. Isothermal means constant temperature. Adiabatic means zero heat exchange.
#10 of 40
ETEA 2023
Permalink
The expression, \(C_p - R\) is equal to:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
1. Concept:Mayer's equation establishes the absolute mathematical relationship between the specific heats of ideal gases.
2. Formula:$$C_p - C_v = R$$
3. Solution:- Start with the standard equation: \(C_p - C_v = R\).
- The question asks for the equivalent of \(C_p - R\).
- By algebraically rearranging the terms (subtracting R from both sides and adding \(C_v\) to both sides), we yield: \(C_p - R = C_v\).
4. Why other options are incorrect:All other options fail simple algebraic rearrangement. \(R\) is the difference itself, not the resulting specific heat.
#11 of 40
ETEA 2023
Permalink
In a certain process, 200J of heat energy is supplied to a system and the same time 50 J of work is done by the system. The increase in the internal energy of the system is:
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
1. Concept:The First Law of Thermodynamics tracks energy conservation. Net energy stored in the system is the energy added minus the energy expelled as work.
2. Formula:$$\Delta U = Q - W$$
3. Solution:- Given heat supplied to the system, \(Q = +200 \text{ J}\).
- Given work done by the system, \(W = +50 \text{ J}\).
- Substitute into the First Law: \(\Delta U = 200 - 50\).
- The resulting increase in internal energy is \(150 \text{ J}\).
4. Why other options are incorrect:Adding 200 and 50 yields 250J (incorrectly assuming work was done ON the system). Other answers result from arbitrary division or ignoring the work completely.
#12 of 40
ETEA 2023
Permalink
Work done during isochoric process is:
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
1. Concept:In thermodynamics, physical work is only accomplished when a boundary is moved against a force, which manifests as a change in volume.
2. Formula:$$W = P \Delta V$$
3. Solution:- An isochoric process strictly means the volume remains perfectly constant throughout the state change.
- Mathematically, this means the change in volume \(\Delta V = 0\).
- Substituting this into the work equation yields \(W = P(0) = 0\).
- Thus, absolutely zero work is done.
4. Why other options are incorrect:Work cannot be positive or negative if the piston physically does not move. Maximum work is typically associated with isobaric processes (for the same volume expansion limits).
#13 of 40
DUHS 2023
Permalink
Which law has a graph similar to the graph of isothermal process?
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
1. Concept:The visual representation (P-V graph) of a thermodynamic process is governed by its mathematical proportionality.
2. Formula:$$PV = \text{constant} \quad \text{(Isothermal)}$$
3. Solution:- An isothermal process operates at a strictly constant temperature.
- Boyle's Law states that at a constant temperature, Pressure and Volume are inversely proportional (\(P \propto \frac{1}{V}\)).
- Both an isothermal process and Boyle's Law mathematically graph as a hyperbola (a downward curving line) on a P-V diagram.
4. Why other options are incorrect:Charles's Law graphs as a straight line (V directly proportional to T). Avogadro's Law relates V and n (straight line). Neither matches the hyperbolic curve of an isothermal expansion.
#14 of 40
DUHS 2023
Permalink
The property of the system that does not change during an adiabatic change
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
1. Concept:An adiabatic process is strictly defined by extreme thermal insulation or extremely rapid expansion/compression, completely preventing heat exchange.
2. Formula:$$\Delta Q = 0$$
3. Solution:- Because the boundary is perfectly insulated, thermal energy (heat) can neither enter nor escape the system.
- Therefore, the total amount of heat (\(Q\)) transferred remains zero; the heat content is unchanged from outside sources.
- In contrast, because work is done, the internal energy, temperature, pressure, and volume all dynamically change.
4. Why other options are incorrect:In an adiabatic compression, work is done ON the gas, so volume decreases, pressure spikes, and temperature rises. Only the heat transfer strictly remains zero.
#15 of 40
DUHS 2023
Permalink
The process in which no external work is performed is called:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
1. Concept:Thermodynamic work is strictly the product of external pressure and the change in the physical boundaries (volume) of the system.
2. Formula:$$W = P \Delta V$$
3. Solution:- For work to be absolute zero, one of the components of the work equation must be zero.
- Since absolute pressure \(P\) of a gas is never zero, the volume change \(\Delta V\) must be zero.
- A process where \(\Delta V = 0\) is termed an Isochoric process (constant volume).
4. Why other options are incorrect:Isothermal, isobaric, and adiabatic processes all permit the volume of the gas to expand or compress, meaning external work is definitively performed in all of them.
#16 of 40
NUMS 2023
Permalink
In isochoric process:
A
Temperature is kept constant
B
Volume is kept constant
C
Exchange of heat is zero
D
Pressure is kept constant
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
1. Concept:Thermodynamic process names reflect the specific physical parameter that is rigidly locked in place during the state change.
2. Formula:$$\Delta V = 0$$
3. Solution:- 'Iso' means constant.
- 'Choric' derives from the Greek 'chora', meaning space or volume.
- Therefore, an Isochoric process is one where the volume of the system is forced to be kept strictly constant, usually by heating a gas in a sealed, rigid steel container.
4. Why other options are incorrect:Constant pressure describes an isobaric process. Constant temperature describes an isothermal process. Zero heat exchange describes an adiabatic process.
#17 of 40
NUMS 2023
Permalink
If 42 J heat is transferred to the system during expansion, what is the change in internal energy when work done is 32 J?
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
1. Concept:The First Law of Thermodynamics operates as an energy balance sheet: Energy In (Heat) minus Energy Out (Work) equals Energy Stored (Internal Energy).
2. Formula:$$\Delta U = Q - W$$
3. Solution:- Heat transferred INTO the system is positive: \(Q = +42 \text{ J}\).
- Work done BY the system during expansion is positive: \(W = +32 \text{ J}\).
- Apply the equation: \(\Delta U = 42 - 32\).
- The net change in internal energy is \(\Delta U = +10 \text{ J}\).
4. Why other options are incorrect:Adding the numbers yields 74 J, which incorrectly assumes work was done ON the system (compression) rather than BY the system (expansion).
#18 of 40
NUMS 2023
Permalink
The \(1^{st}\) law of thermodynamic is the generalization of the law of conservation of:
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
1. Concept:The First Law of Thermodynamics bridges the gap between mechanics and thermal physics by proving that heat is simply another transferable form of energy.
2. Formula:$$\Delta U = Q - W$$
3. Solution:- Before this law, heat and mechanical work were thought to be completely separate entities.
- The First Law dictates that the total energy (accounting for both heat and work) in an isolated universe remains strictly constant.
- Thus, it is the direct thermal generalization of the universal Law of Conservation of Energy.
4. Why other options are incorrect:Conservation of momentum applies strictly to forces and collisions. Conservation of charge applies to electromagnetism. Thermodynamics inherently deals with energy.
#19 of 40
UHS 2022
Permalink
Temperature of given mass of a gas is changed from \(150^\circ \text{C}\) to \(300^\circ \text{C}\) during an isobaric process, volume of the gas will become:
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
1. Concept:Charles's Law dictates that at constant pressure, volume is directly proportional to absolute temperature (Kelvin), NOT Celsius.
2. Formula:$$\frac{V_1}{T_1} = \frac{V_2}{T_2}$$
3. Solution:- Convert initial temperature to Kelvin: \(T_1 = 150 + 273 = 423 \text{ K}\).
- Convert final temperature to Kelvin: \(T_2 = 300 + 273 = 573 \text{ K}\).
- The ratio \(\frac{T_2}{T_1} = \frac{573}{423} \approx 1.35\).
- Since \(V_2 = 1.35 V_1\), the volume increases but does not double. It is strictly less than double.
4. Why other options are incorrect:A common student error is to look at \(150^\circ \text{C}\) and \(300^\circ \text{C}\) and assume the temperature doubled. Thermodynamics requires all ratios to be computed in absolute Kelvin.
#20 of 40
SZABMU 2022
Permalink
The thermodynamic process during which volume of the system remains constant is called:
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
1. Concept:Thermodynamic processes are named based on the property that remains rigid or unchanging throughout the physical transformation.
2. Formula:$$\Delta V = 0 \implies W = 0$$
3. Solution:- The prefix 'iso-' means equal or constant.
- The suffix '-choric' pertains to space or volume.
- Therefore, an Isochoric process is defined strictly as a thermodynamic change where the container's volume remains completely unchanged.
4. Why other options are incorrect:Isobaric means constant pressure. Isothermal means constant temperature. Adiabatic means zero heat transfer.
#21 of 40
SZABMU 2022
Permalink
The conditions for application of Boyle's law holds good in:
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
1. Concept:Boyle's Law states that the pressure of a given mass of an ideal gas is inversely proportional to its volume, provided the temperature remains absolutely constant.
2. Formula:$$P \propto \frac{1}{V} \quad (\text{at constant } T)$$
3. Solution:- Because Boyle's law strictly requires a constant temperature to be mathematically valid (\(PV = \text{constant}\)), the thermodynamic environment must enforce this.
- A process that maintains a constant temperature is defined as an isothermal process.
4. Why other options are incorrect:Isobaric processes enforce constant pressure (Charles's Law). Isochoric processes enforce constant volume (Gay-Lussac's Law). Adiabatic processes allow temperature to fluctuate.
#22 of 40
SZABMU 2022
Permalink
The internal energy of a system during an isothermal process:
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
1. Concept:For an ideal gas, internal energy is purely a function of its absolute kinetic temperature. If the temperature does not change, the internal kinetic energy cannot change.
2. Formula:$$\Delta U = n C_v \Delta T$$
3. Solution:- An isothermal process is characterized by \(\Delta T = 0\).
- Plugging this into the internal energy equation yields \(\Delta U = 0\).
- Because the change is zero, the total internal energy remains constant throughout the entire process.
4. Why other options are incorrect:Internal energy cannot decrease or increase unless the temperature changes. It cannot become zero unless the gas reaches absolute zero (0 K), which is not the case for a standard isothermal process.
#23 of 40
SZABMU 2022
Permalink
Two bodies A and B of temperature \(T_A = 100^\circ \text{C}\) and \(T_B = 0^\circ \text{C}\) are brought in thermal contact with each other. Which one in the followings is possible at thermal equilibrium?
A
\(T_A = 0^\circ \text{C}, T_B = 100^\circ \text{C}\)
B
\(T_A = 60^\circ \text{C}, T_B = 50^\circ \text{C}\)
C
\(T_A = 60^\circ \text{C}, T_B = 40^\circ \text{C}\)
D
\(T_A = 45^\circ \text{C}, T_B = 45^\circ \text{C}\)
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
1. Concept:According to the Zeroth Law of Thermodynamics, two systems in thermal contact will exchange heat until they reach thermal equilibrium, at which point their temperatures must be exactly equal.
2. Formula:$$T_A = T_B = T_{\text{equilibrium}}$$
3. Solution:- Heat will flow from the hotter body (A) to the colder body (B).
- This flow continues until the temperatures equalize.
- Looking at the options, the only mathematical state where \(T_A\) exactly equals \(T_B\) is \(45^\circ \text{C}\). The specific final value depends on mass and specific heat, but equality is strictly required.
4. Why other options are incorrect:Options with unequal temperatures signify that thermal equilibrium has not yet been reached and heat is still flowing.
#24 of 40
ETEA 2022
Permalink
In an isolated thermodynamic system:
A
No dissipated energy and heat are transferred to the environment
B
Neither heat nor any mass are transferred to the environment
C
No mass transfers to the environment
D
No heat transfers to the environment
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
1. Concept:Thermodynamic systems are classified by their boundaries. An 'isolated' system represents the most restrictive boundary possible in physics.
2. Formula:$$\Delta m = 0 \quad \text{and} \quad Q = 0, W = 0$$
3. Solution:- An OPEN system allows both mass and energy to cross its boundaries.
- A CLOSED system allows energy (heat/work) to cross, but mass is blocked.
- An ISOLATED system is completely sealed: absolutely no mass can escape, and absolutely no energy (neither heat nor work) can cross the boundary.
4. Why other options are incorrect:Saying 'no heat transfers' only describes an adiabatic or closed system. Saying 'no mass transfers' only describes a closed system. Only 'neither heat nor mass' accurately defines total isolation.
#25 of 40
BUMHS 2022
Permalink
Thermal equilibrium in environment occurs due to:
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
1. Concept:Thermal equilibrium is achieved when two regions in thermal contact cease to have a net exchange of thermal energy, meaning they have reached the exact same temperature.
2. Formula:$$\Delta Q = mc \Delta T$$
3. Solution:- If there is a temperature gradient in an environment, nature seeks balance.
- This balance is achieved through the physical flow of heat from the higher temperature region to the lower temperature region.
- Therefore, it is the fundamental mechanism of 'heat flow' that eventually neutralizes the gradient, resulting in equilibrium.
4. Why other options are incorrect:Heat generation actively disrupts equilibrium. Absorption and dissipation are singular viewpoints of the mechanism; 'heat flow' encompasses the entire interactive transfer process.
#26 of 40
DUHS 2022
Permalink
Specific heat capacity of a gas is measured in:
A
\(\text{J kg}^{-1} \text{K}^{-1}\)
C
\(\text{J kg}^{-1\circ} \text{K}\)
D
\(\text{J kg}^\circ \text{K}\)
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
1. Concept:Specific heat capacity determines how much thermal energy is required to raise the temperature of one unit of mass by one unit of absolute temperature.
2. Formula:$$c = \frac{Q}{m \Delta T}$$
3. Solution:- Look at the units in the formula: \(Q\) is in Joules (J).
- Mass \(m\) is in kilograms (kg).
- Temperature change \(\Delta T\) is in Kelvin (K).
- Combining these gives: \(\frac{\text{J}}{\text{kg} \cdot \text{K}}\), which is mathematically written as \(\text{J kg}^{-1} \text{K}^{-1}\).
4. Why other options are incorrect:Other options have incorrect inverse signs (e.g., placing mass in the numerator rather than denominator) or use inappropriate degrees for Kelvin.
#27 of 40
DUHS 2022
Permalink
First law of thermodynamics is mathematically represented as:
A
\(\Delta U = \Delta Q - \Delta W\)
B
\(\Delta Q = \Delta W - \Delta U\)
C
\(\Delta U = \Delta W - \Delta Q\)
D
\(\Delta W = \Delta Q + \Delta U\)
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
1. Concept:The First Law of Thermodynamics is the law of conservation of energy applied to thermal systems, balancing heat, internal energy, and work.
2. Formula:$$\Delta Q = \Delta U + \Delta W$$
3. Solution:- The standard sign convention equation states that heat added to a system (\(\Delta Q\)) causes a change in internal energy (\(\Delta U\)) and does work on the surroundings (\(\Delta W\)).
- Rearranging this algebraically to isolate internal energy gives: \(\Delta U = \Delta Q - \Delta W\).
4. Why other options are incorrect:The other equations fail basic algebraic rearrangement of the First Law or imply incorrect physical phenomena (like work generating heat without changing internal energy).
#28 of 40
NUMS 2022
Permalink
First law of thermodynamic is based on:
A
Law of conservation of charge
B
Law of conservation of energy
C
Law of conservation of mass
D
Law of conservation of momentum
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
1. Concept:The First Law of Thermodynamics establishes that energy can be transformed from one form to another, but can be neither created nor destroyed.
2. Formula:$$\Delta Q = \Delta U + \Delta W$$
3. Solution:- The equation states that whatever thermal energy is added to the system must exactly equal the energy stored within the molecules plus the mechanical energy expended as work.
- This is a direct, macroscopic manifestation of the universal Principle of Conservation of Energy.
4. Why other options are incorrect:Thermodynamics does not strictly govern particle charge, mass creation (outside nuclear physics), or mechanical momentum in this specific context.
#29 of 40
NUMS 2022
Permalink
The thermodynamics process during which the pressure is kept constant is called:
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
1. Concept:Different thermodynamic processes are characterized by the state variable (P, V, T, Q) that is artificially or naturally constrained to remain unchanged.
2. Formula:$$\Delta P = 0$$
3. Solution:- The prefix 'iso-' means equal or unchanged.
- The root 'baric' relates to pressure (e.g., barometer, bars).
- Therefore, a process strictly operating under a fixed, constant pressure is termed an Isobaric process.
4. Why other options are incorrect:Isochoric locks volume. Isothermal locks temperature. Adiabatic locks heat transfer. None of these guarantee constant pressure.
#30 of 40
PMDC 2021
Permalink
\(C_p - C_v\) value, find if 3 moles of gas is given
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
1. Concept:The difference between the total heat capacity at constant pressure and constant volume for 'n' moles of an ideal gas equals \(nR\).
2. Formula:$$C_p - C_v = nR$$
3. Solution:- The standard molar specific heat relation is \(c_p - c_v = R\) (for 1 mole).
- The question refers to the total heat capacities \(C_p\) and \(C_v\) for 3 moles of the gas.
- By substituting \(n = 3\) into the extensive property relation, we get \(C_p - C_v = 3R\).
4. Why other options are incorrect:Options like 2R, 5R, or 7R are purely distractor values and do not map to the number of moles provided in the question stem.
#31 of 40
PMDC 2020
Permalink
In which process the entire of heat supplied to the gas is converted to the internal energy of the gas?
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
1. Concept:If a gas is heated in a rigid container, its volume cannot change. Since work done is pressure multiplied by the change in volume, zero volume change means zero work done.
2. Formula:$$Q = \Delta U + W$$
3. Solution:- In an isochoric process, volume is constant (\(\Delta V = 0\)).
- Work done \(W = P \Delta V = 0\).
- Applying the First Law of Thermodynamics: \(Q = \Delta U + 0\).
- Therefore, all heat supplied (\(Q\)) increases the internal energy (\(\Delta U\)).
4. Why other options are incorrect:In isothermal, all heat converts to work. In isobaric, heat splits into internal energy and work. In adiabatic, no heat is supplied at all.
#32 of 40
PMDC 2020
Permalink
The internal energy of a system during an isothermal process:
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
1. Concept:For an ideal gas, the internal energy is directly proportional to its absolute temperature.
2. Formula:$$U \propto T \implies \Delta U = n C_v \Delta T$$
3. Solution:- An isothermal process is defined as a thermodynamic change where the temperature remains constant (\(\Delta T = 0\)).
- Since the change in temperature is zero, the change in internal energy \(\Delta U\) is also zero.
- Hence, the internal energy remains completely constant throughout the process.
4. Why other options are incorrect:Internal energy would increase during heating (temp rise) and decrease during cooling (temp fall), but cannot change if temperature is locked.
#33 of 40
NUMS 2020
Permalink
In a certain process, 400J of heat energy is supplied to a system and at the same time 150J of work is done by the system. The increase in internal energy of system is
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
1. Concept:The First Law of Thermodynamics dictates that energy is conserved. The heat added to a system splits into changing its internal energy and doing work on the surroundings.
2. Formula:$$Q = \Delta U + W$$
3. Solution:- Identify given values with correct sign conventions: Heat supplied \(Q = +400 \text{ J}\), Work done BY the system \(W = +150 \text{ J}\).
- Rearrange the formula to solve for internal energy: \(\Delta U = Q - W\).
- Substitute values: \(\Delta U = 400 - 150\).
- Result: \(\Delta U = 250 \text{ J}\).
4. Why other options are incorrect:Adding the numbers instead of subtracting gives 550J. Failing to account for work gives 400J. The correct energy balance requires subtracting the energy lost as work.
#34 of 40
NUMS 2020
Permalink
The rapid escape of air from a burst tyre is an example of:
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
1. Concept:Thermodynamic processes that happen extremely fast do not leave enough time for heat to exchange with the surrounding environment.
2. Formula:$$Q = 0 \implies W = -\Delta U$$
3. Solution:- When a tire bursts, the high-pressure air expands rapidly into the atmosphere.
- Because the expansion is highly sudden, there is practically zero time for heat to flow into or out of the gas (\(Q = 0\)).
- Any process where heat exchange is zero is strictly defined as an adiabatic process.
4. Why other options are incorrect:Isothermal processes must happen very slowly to allow heat exchange. Isochoric processes require a constant volume. Isobaric requires constant pressure.
#35 of 40
MDCAT 2019
Permalink
The sum of all forms of molecular energies (kinetic and potential) of a substance is termed as?
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
1. Concept:In thermodynamics, the macroscopic properties of a system are defined by the microscopic energies of its particles. The total microscopic energy is called Internal Energy.
2. Formula:$$U = \Sigma E_k + \Sigma E_p$$
3. Solution:- By definition, the internal energy (\(U\)) of a thermodynamic system is the sum of the random kinetic energies of its molecules and the potential energies associated with intermolecular forces.
4. Why other options are incorrect:Heat is energy in transit, not energy contained within a system. Absolute energy and elastic energy do not properly describe the sum of all microscopic molecular energies.
#36 of 40
MDCAT 2018
Permalink
If one mole of an ideal gas is heated at constant pressure, then the first law of thermodynamics can be written as:
A
\(C_v \Delta T = C_p \Delta T + P \Delta T\)
B
\(C_p \Delta T = C_v \Delta T + V \Delta P\)
C
\(C_p \Delta T = C_v \Delta T + P \Delta V\)
D
\(\Delta C_v T = \Delta C_p T + P \Delta V\)
View Answer & Propolis Autopsy
Correct Key: Option C
Diagnostic Explanation
1. Concept:The First Law of Thermodynamics states that heat supplied to a system equals the increase in internal energy plus the work done by the system.
2. Formula:$$Q = \Delta U + W$$
3. Solution:- At constant pressure, heat supplied for one mole is \(Q = C_p \Delta T\).
- The change in internal energy for any process is \(\Delta U = C_v \Delta T\).
- The work done at constant pressure is \(W = P \Delta V\).
- Substituting these yields: \(C_p \Delta T = C_v \Delta T + P \Delta V\).
4. Why other options are incorrect:Other options misrepresent the basic thermodynamic terms, swapping specific heats or introducing non-standard terms like \(V \Delta P\) which applies to steady-flow processes rather than constant pressure expansion.
#37 of 40
MDCAT 2018
Permalink
If \(C_v = \frac{5}{2} R\), \(C_p\) will be
View Answer & Propolis Autopsy
Correct Key: Option B
Diagnostic Explanation
1. Concept:Molar specific heat at constant pressure (\(C_p\)) is always greater than molar specific heat at constant volume (\(C_v\)) by an amount equal to the universal gas constant (\(R\)).
2. Formula:$$C_p - C_v = R$$
3. Solution:- Rearrange the relation: \(C_p = C_v + R\).
- Substitute the given value: \(C_p = \frac{5}{2} R + R\).
- Calculate the sum: \(C_p = \frac{5}{2} R + \frac{2}{2} R = \frac{7}{2} R\).
4. Why other options are incorrect:Other options result from incorrect arithmetic or by erroneously subtracting \(R\) instead of adding it.
#38 of 40
ETEA 2018
Permalink
The amount of heat required to raise the temperature of 10 moles of water from 70K to 80K (molar heat capacity of water 75.24J) is:
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
1. Concept:The heat required to raise the temperature of a given number of moles of a substance depends on its molar heat capacity and the change in temperature.
2. Formula:$$Q = n C \Delta T$$
3. Solution:- Identify the given values: \(n = 10 \text{ moles}\), \(C = 75.24 \text{ J/mol K}\).
- Calculate temperature change: \(\Delta T = 80 \text{ K} - 70 \text{ K} = 10 \text{ K}\).
- Substitute into the formula: \(Q = 10 \times 75.24 \times 10\).
- Calculate final value: \(Q = 100 \times 75.24 = 7524 \text{ J}\).
4. Why other options are incorrect:The other options represent decimal errors (forgetting to multiply by the 10 moles or the 10 K temperature difference) or simple calculation mistakes.
#39 of 40
ETEA 2014
Permalink
The amount of heat required to raise the temperature of 1kg of substance through 1 K is called;
View Answer & Propolis Autopsy
Correct Key: Option A
Diagnostic Explanation
1. Concept:The specific heat capacity of a material is defined as the amount of heat energy required to raise the temperature of a unit mass (1 kg in SI units) of that substance by 1 Kelvin (or 1°C).
2. Formula:$$c = \frac{Q}{m \Delta T}$$
3. Solution:- The question explicitly specifies a mass of 1 kg.
- When mass \(m = 1 \text{ kg}\) and \(\Delta T = 1 \text{ K}\), the heat \(Q\) is exactly equal to the specific heat \(c\).
4. Why other options are incorrect:Heat capacity (without the word 'specific') refers to the entire body regardless of its mass. Joules and calories are units of energy, not properties of a material.
#40 of 40
MDCAT 2013
Permalink
What is the factor upon which change in internal energy of an ideal gas depends?
B
Path followed to change internal energy
C
Changed in volume and temperature
View Answer & Propolis Autopsy
Correct Key: Option D
Diagnostic Explanation
1. Concept:The internal energy of an ideal gas is purely translational kinetic energy, which depends exclusively on the absolute temperature of the gas.
2. Formula:$$U = \frac{f}{2} n R T$$
3. Solution:- Since the number of moles \(n\) and the gas constant \(R\) are constant for a fixed system, any change in internal energy \(\Delta U\) is directly proportional to the change in temperature \(\Delta T\).
- Therefore, it does not depend on the path or volume changes.
4. Why other options are incorrect:Internal energy is a state function, meaning it is completely independent of the path taken. For an ideal gas specifically, intermolecular forces are zero, so volume changes do not alter potential energy.
The Complete Combat Engine
Ready to Score 180+ in MDCAT 2026?
Join thousands of pre-med students utilizing BeambePrep's full combat suite: Swarm Mode timed challenges, Normal Grind Mode, Propolis Ward mistake notebooks, Infinite Full-Length Practice (FLP) Mocks, and FSRS Spaced Repetition.