Question

The value of $C_{p} - C_{v} = 1.00R$ for a gas in state A and $C_{p} - C_{v} = 1.06R$ 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)

$P_{A} < P_{B}$, $T_{A} > T_{B}$

Answer 2

For state A, $C_{p} - C_{v} = R$ i.e. the gas behaves as ideal gas.

For state B, $C_{p} - C_{v} = 1.06R( \neq R)$ i.e. the gas does not behave like ideal gas.

and 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 $T_{A} > T_{B}$