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Thermodynamics question

2018 · 15 Apr · Shift 1 · Q10
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Thermodynamics question

2018 · 15 Apr · Shift 1 · Q10

JEE MainChemistryThermodynamicsMCQ+4 / −1
An ideal gas undergoes a cyclic process as shown in Figure. JEE Main 2018 (Online) 15th April Morning Slot Chemistry - Thermodynamics Question 174 English Δ\DeltaΔ UBC = −-− 5 kJ mol-1, qAB = 222 kJ mol-1, WAB =−-− 5 kJ mol-1, WCA = 3 kJ mol-1. Heat absorbed by the system during process CACACA is :
  1. A
    −-− 5 kJ mol-1
  2. B
    +5 kJ mol-1
  3. C
    18 kJ mol-1
  4. D
    −-− 18 kJ mol-1
View written solutionFree

Correct answer: B

  1. Use the first law of thermodynamics

For a process,

ΔU=q+W\Delta U = q + WΔU=q+W

where here the given sign convention is consistent with:

  • q>0q>0q>0 : heat absorbed by system
  • W>0W>0W>0 : work done on the system

This is clear from the data for process ABABAB.

  1. Find ΔUAB\Delta U_{AB}ΔUAB​

Given:

qAB=2 kJ mol−1,WAB=−5 kJ mol−1q_{AB}=2\ \text{kJ mol}^{-1}, \qquad W_{AB}=-5\ \text{kJ mol}^{-1}qAB​=2 kJ mol−1,WAB​=−5 kJ mol−1

So,

ΔUAB=qAB+WAB=2+(−5)=−3 kJ mol−1\Delta U_{AB}=q_{AB}+W_{AB}=2+(-5)=-3\ \text{kJ mol}^{-1}ΔUAB​=qAB​+WAB​=2+(−5)=−3 kJ mol−1
  1. Use cyclic condition to find ΔUCA\Delta U_{CA}ΔUCA​

For a cyclic process,

ΔUAB+ΔUBC+ΔUCA=0\Delta U_{AB}+\Delta U_{BC}+\Delta U_{CA}=0ΔUAB​+ΔUBC​+ΔUCA​=0

Given:

ΔUBC=−5 kJ mol−1\Delta U_{BC}=-5\ \text{kJ mol}^{-1}ΔUBC​=−5 kJ mol−1

Thus,

−3+(−5)+ΔUCA=0-3+(-5)+\Delta U_{CA}=0−3+(−5)+ΔUCA​=0 ΔUCA=8 kJ mol−1\Delta U_{CA}=8\ \text{kJ mol}^{-1}ΔUCA​=8 kJ mol−1
  1. Find heat absorbed during process CACACA

Again using

ΔUCA=qCA+WCA\Delta U_{CA}=q_{CA}+W_{CA}ΔUCA​=qCA​+WCA​

Given:

WCA=3 kJ mol−1W_{CA}=3\ \text{kJ mol}^{-1}WCA​=3 kJ mol−1

So,

8=qCA+38=q_{CA}+38=qCA​+3 qCA=5 kJ mol−1q_{CA}=5\ \text{kJ mol}^{-1}qCA​=5 kJ mol−1
  1. Final answer

Heat absorbed during process CACACA is

+5 kJ mol−1\boxed{+5\ \text{kJ mol}^{-1}}+5 kJ mol−1​

So the correct option is B.

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