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# A perfect gas goes from state A to another state B by absorbing 8×105J of heat and doing 6.5×105J of external work. It is now transferred between the same two states in another process in which it absorbs 105J of heat. In the second process

A
work done by the gas is 105J
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B
work done on gas is 105J
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C
work done by gas is 0.5×105J
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D
work done on the gas is 0.5×105J
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Solution

## The correct option is D work done on the gas is 0.5×105JdU=dQ−dW=(8×105−6.5×105)=1.5×105JdW=dQ−dU=105−1.5×105=−0.5×105J- ve sign indicates that work done on the gas is 0.5×105J

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