For a gas,C_{p} - C_{V} = R in state P and C_ | vedclass

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(D) For an ideal gas,the relation $C_{p} - C_{V} = R$ holds true.Real gases behave as ideal gases at high temperatures and low pressures.In state $P$,

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$T_{P} > T_{Q}$

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(D) For an ideal gas,the relation $C_{p} - C_{V} = R$ holds true.Real gases behave as ideal gases at high temperatures and low pressures.In state $P$,$C_{p} - C_{V} = R$,which indicates that the gas is behaving as an ideal gas.In state $Q$,$C_{p} - C_{V} = 1.10 R$,which is greater than $R$. This indicates that the gas is deviating from ideal behavior,which typically happens at lower temperatures.Since state $P$ represents ideal behavior and state $Q$ represents non-ideal behavior,it implies that the temperature in state $P$ must be higher than in state $Q$.Therefore,$T_{P} > T_{Q}$.

For a gas,$C_{p} - C_{V} = R$ in state $P$ and $C_{p} - C_{V} = 1.10 R$ in state $Q$. If $T_{P}$ and $T_{Q}$ are the temperatures in states $P$ and $Q$ respectively,then which of the following is true?

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