The value of C_p - C_v = 1.00 R for a gas in | vedclass

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(C) For an ideal gas,the Mayer's relation is given by $C_p - C_v = R$.In state $A$,$C_p - C_v = 1.00 R$,which implies the gas behaves as an ideal gas

Vedclass · Accepted Answer

$P_A  T_B$

Vedclass · Answer

(C) For an ideal gas,the Mayer's relation is given by $C_p - C_v = R$.In state $A$,$C_p - C_v = 1.00 R$,which implies the gas behaves as an ideal gas in this state.In state $B$,$C_p - C_v = 1.06 R$,which is not equal to $R$,implying the gas deviates from ideal behavior in this state.Real gases approach ideal behavior at high temperatures and low pressures.Since state $A$ is closer to ideal behavior than state $B$,it must be at a higher temperature and lower pressure compared to state $B$.Therefore,$P_A  T_B$.

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 $T_B$ denote the temperature in the two states,then:

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