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Standard IX

Chemistry

The Ideal Gas Equation

The value of ...

Question

The value of $C_{p} - C_{v}$ is $1.09 R$ for a gas sample in state $A$ and is $1.00 R$ in state $B$ . Let $T_{A}, T_{B}$ denote the temperature and $P_{A}$ and $P_{B}$ denote the pressure of the states $A$ and $B$ respectively. Then:

P_{A} < P_{B}

and

T_{A} > T_{B}

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P_{A} > P_{B}

and

T_{A} > T_{B}

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P_{A} = P_{B}

and

T_{A} = T_{B}

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P_{A} > P_{B}

and

T_{A} < T_{B}

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Solution

The correct option is A $P_{A} < P_{B}$ and $T_{A} > T_{B}$
For state A, $C_{P} - C_{V} = R$ i.e. the gas behaves as ideal gas.
For state B, $C_{P} - C_{V} = 1.06 R (\neq R)$ i.e. the gas does not behave like ideal gas and we know that at high temperature & at low pressure the nature of gas may be ideal.
So, answer is option $A$ .

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Figure shows two vessels A and B with rigid walls containing ideal gases. The pressure, temperature and the volume are $p_{A}, T_{A}$ , V in the vessel A and $p_{B}, T_{B}$ , V in the vessel B. The vessels are now connected through a small tube. Show that the pressure p and the temperature T satisfy $\frac{p}{T} = \frac{1}{2} (\frac{p_{A}}{T_{A}} + \frac{p_{B}}{T_{B}})$ when equilibrium is achieved.

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The value of Cp−Cv is 1.09R for a gas sample in state A and is 1.00R in state B. Let TA,TB denote the temperature and PA and PB denote the pressure of the states A and B respectively. Then:

The correct option is A PA<PB and TA>TBFor state A, CP−CV=R i.e. the gas behaves as ideal gas.For state B, CP−CV=1.06R(≠R) i.e. the gas does not behave like ideal gas and we know that at high temperature & at low pressure the nature of gas may be ideal.So, answer is option A.

The value of $C_{p} - C_{v}$ is $1.09 R$ for a gas sample in state $A$ and is $1.00 R$ in state $B$ . Let $T_{A}, T_{B}$ denote the temperature and $P_{A}$ and $P_{B}$ denote the pressure of the states $A$ and $B$ respectively. Then: