Question

The value of $C_{p}-C_{v}=1.00\, R$ for a gas in state $A$ and $C_{p}-C_{v}=1.06 \, R$ in another state. If $P_{A}$ and $P_{B}$ denote the pressure and $T_{A}$ and $T_{B}$ denote the temperatures in the two states, then

• A. $P_{A}=P_{B}, T_{A}>T_{B}$
• B. $P_{A}>P_{B}, T_{A}=T_{B}$
• C. $P_{A} < P_{B}, T_{A}>T_{B}$
• D. $P_{A}=P_{B}, T_{A} < T_{B}$

Answer 1

Answer: (C)

For state $A, C_{p}-C_{v}=R$, i.e., the gas behaves as an ideal gas. For state $B, C_{p}-C_{v}=1.06 \, R(\neq R)$, i.e., the gas does not behave like an ideal gas.
We know that at high temperature and at low pressure, nature of gas may be ideal.
So we can say that $P_{A}< P_{B}$ and $T_{A}>T_{B}$.