Chemistry Thermochemistry ETEA 2024
PMDC Verified Question 11 of 81
When 1 mole of ice melts at \( 0^\circ\text{C} \) and constant pressure of 1 atmosphere, 6025 J of heat is absorbed by the system. The molar volume of ice and water are 0.020 and \( 0.018 \text{ dm}^3 \), respectively. Calculate \( \Delta E \). (\( 1 \text{ dm}^3\text{.atm} = 101.33 \text{ J} \))
A
\( 6010.20 \text{ J} \)
B
\( 6020.20 \text{ J} \)
C
\( 6015.20 \text{ J} \)
D
\( 6025.20 \text{ J} \)
Tap any option to test your recall and reveal the step-by-step Propolis autopsy.

Propolis Cognitive Error Autopsy

Official Correct Choice:
Option D: \( 6025.20 \text{ J} \)
Concept:

The first law of thermodynamics requires accounting for the work done when a system changes volume, even during phase changes like melting.

Formula:

$$ \Delta E = q - P\Delta V $$

Solution:

  • Heat absorbed (\( q \)) = \( +6025 \text{ J} \).


  • Change in volume (\( \Delta V \)) = \( V_{\text{water(liquid)}} - V_{\text{ice(solid)}} \).
    \( \Delta V = 0.018 - 0.020 = -0.002 \text{ dm}^3 \) (Water uniquely shrinks when melting).


  • Calculate Work (\( W = P\Delta V \)) in Joules:
    \( W = 1 \text{ atm} \times -0.002 \text{ dm}^3 = -0.002 \text{ atm.dm}^3 \).
    Convert to Joules: \( -0.002 \times 101.33 = -0.2026 \text{ J} \).


  • Calculate Internal Energy:\( \Delta E = 6025 - (-0.2026) = 6025.2026 \text{ J} \).


  • Rounded to two decimal places, this matches exactly \( 6025.20 \text{ J} \).


Why other options are incorrect:

Other options result from incorrectly adding the volume change as expansion (which would subtract work) or ignoring the contraction physics entirely.

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