Question

In the process $P{V}^2$ = constant, if temperature of gas is increased

• A. change in the internal energy of gas is positive
• B. work done by gas is positive
• C. heat is given to the gas
• D. heat is taken out from the gas

Answer 1

Answer: (A), (C)

The correct options are
A change in the internal energy of gas is positive
C heat is given to the gas

given process is a polytropic process,so from the first law of thermodynamics we have
$dQ=dU+dW$
now $U=C_{v}dT$ and it is mention that temperature of gas is increased,so

$\Delta U=U_f-U_i$ hence $U_f$ is increases so it will always be positive.

$W=\frac{P_1{V}_1-P_2{V}_2}{n-1}$ so it will be negative
now from first law of thermodynamics if internal energy is positive then $Q$ will always be positive ,so we can say that heat is given to the system.