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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 $B$ . If $P_A$ and $P_B$ denote the pressure and $T_A$ and $T_B$ denotes the temperature in the two states, then
$P_A = P_B\, ;\, T_A > T_B$
$P_A > P_B\, ;\, T_A = T_B$
$P_A < P_B\, ;\, T_A > T_B$
$P_A = P_B\, ;\, T_A < T_B$
Solution
For state $\mathrm{A}, \mathrm{C}_{\mathrm{D}}-\mathrm{C}_{\mathrm{v}}=\mathrm{R}$ i.e. the gas behaves as ideal gas.
For state $\mathrm{B}, \mathrm{C}_{\mathrm{p}}-\mathrm{C}_{\mathrm{v}}=1.06 \mathrm{R}(\neq \mathrm{R})$ 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 $(c)$ $\mathrm{P}_{\mathrm{A}}<\mathrm{P}_{\mathrm{B}}$ and $\mathrm{T}_{\mathrm{A}}>\mathrm{T}_{\mathrm{B}}$
