| [a] Heat required in the process is 30 K Cal |
| [b] Increase in the internal energy of gas is 21 Kcal |
| [c] If process is at constant volume then heat required is 21 Kcal |
| [d] external work done is 10 KCal |
A) a, d
B) a, b, c
C) c, d
D) b, d
Correct Answer: B
Solution :
[a] Heat required is- \[Q=nCp\Delta T=n\left( Cv+R \right)\Delta T\] \[=\,\,\frac{1000}{28}[5+2]\times 120=30\times {{10}^{3}}cal\] \[Q=30\text{ }Kcal\] [b] The increase in the internal energy is \[\Delta U\,\,=\,\,nCv\Delta T=\frac{1000}{28}\,\,\times 5\times 120\] \[=\text{ }21\text{ }Kcal\] [c] \[Q=nCv\Delta T+P\Delta V\] for constant volume \[\Delta V=0\] \[Q=nCv\Delta T=21\,Kcal\] external work done is \[W=Q-\Delta U=8.6\,\,KCal\]
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