Thermodynamics Question 306
Question: An ideal gas is subjected to cyclic process involving four thermodynamic states, the amounts of heat (Q) and work (W) involved in each of these states $ Q _{1}=6000J;Q _{2}=-5500J;Q _{3}=-3000J $ $ Q _{4}=+3500J $$ W _{1}=2500J;W _{2}=-1000J;W _{3}=-1200J $$ W _{4}=xJ $ The ratio of the net work done by the gas to the total heat absorbed by the gas is $ \eta $ . The values of x and n respectively are
Options:
A) 500; 7.5%
B) 700; 10.5%
C) 1000; 21%
D) 1500; 15%
Show Answer
Answer:
Correct Answer: B
Solution:
[b] $ Q=Q _{1}+Q _{2}+Q _{3}+Q _{4} $
$ =6000-5500-3000+3500=+1000J $
$ W=W _{1}+W _{2}+W _{3}+W _{4} $
$ =2500-1000-1200+x=+300+x $ In cyclic process, $ \Delta U=0 $ Now, $ Q=\Delta U+W $
$ \text{or }1000=0+( 300+x ) $
$ \therefore x=700J $
$ \eta =\frac{W}{Q _{1}+Q _{4}} $
$ =\frac{1000}{6000+3500}=10.5% $
BETA
