Thermodynamics Question 142
Question: A perfect gas goes from state A to another state B by absorbing $ 8\times 10^{5}J $ of heat and doing $ 6.5\times 10^{5}J $ of external work. It is now transferred between the same two states in another process in which it absorbs $ 10^{5}J $ of heat. Then in the second process
[BHU 1997]
Options:
A) Work done on the gas is $ 0.5\times 10^{5}J $
B) Work done by gas is $ 0.5\times 10^{5}J $
C) Work done on gas is $ 10^{5}J $
D) Work done by gas is $ 10^{5}J $
Show Answer
Answer:
Correct Answer: A
Solution:
In first process using $ \Delta Q=\Delta U+\Delta W $
Therefore $ 8\times 10^{5}=\Delta U+6.5\times 10^{5} $
Therefore $ \Delta U=1.5\times 10J $
Since final and initial states are same in both process
So $ \Delta U $ will be same in both process
For second process using $ \Delta Q=\Delta U+\Delta W $
Therefore $ 10^{5}=1.5\times 10^{5}+\Delta W $
Therefore $ \Delta W=-0.5\times 10^{5}J $
BETA
