Physics Thermodynamics UHS 2022
PMDC Verified Question 33 of 54
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:
A
Double
B
Half
C
Remains same
D
Less than double
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option D: Less than double
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.

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