A) internal energy of the system decreases
B) infernal energy of the system increases
C) work done by the gas is negative
D) work done by the gas is positive
Correct Answer: C
Solution :
In an isothermal process, work done is \[dW=pdV\] \[\therefore \] \[dW=p({{V}_{2}}-{{V}_{1}})\] \[=p\left( \frac{V}{2}-V \right)\] \[\left( \because {{V}_{1}}=V,{{V}_{2}}=\frac{V}{2} \right)\] \[=-\frac{pV}{2}\] So, the work done by the gas is negative.
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