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 $



NCERT Chapter Video Solution

Dual Pane