The value of Cp-Cv=1.00 R for a gas in state A and Cp-Cv=1.06 R in another state. If PA and PB denote the pressure and TA and TB denote the temperatures in the two states, then

Question

The value of Cp-Cv=1.00 R for a gas in state A and Cp-Cv=1.06 R in another state. If PA and PB denote the pressure and TA and TB denote the temperatures in the two states, then (A) PA=PB, TA>TB (B) PA>PB, TA=TB (C) PA < PB, TA>TB (D) PA=PB, TA < TB

Tardigrade Admin · Accepted Answer

For state A, Cp-Cv=R, i.e., the gas behaves as an ideal gas. For state B, Cp-Cv=1.06 R(≠ 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 PA< PB and TA>TB.

Q. 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

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

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

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

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

Q. The value of Cp​−Cv​=1.00R for a gas in state A and Cp​−Cv​=1.06R in another state. If PA​ and PB​ denote the pressure and TA​ and TB​ denote the temperatures in the two states, then

PA​=PB​,TA​>TB​

PA​>PB​,TA​=TB​

PA​<PB​,TA​>TB​

PA​=PB​,TA​<TB​

For state A,Cp​−Cv​=R, i.e., the gas behaves as an ideal gas. For state B,Cp​−Cv​=1.06R(=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 PA​<PB​ and TA​>TB​.

Q. 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

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

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

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

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