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Standard XII

Physics

Calorimetry

The volume ...

Question

The volume $V$ versus temperature $T$ graphs for a certain amount of a perfect gas at two pressure P $_{1}$ and P $_{2}$ are as shown in figure. Then

P_{1} = P_{2}

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P_{1} > P_{2}

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P_{1} < P_{2}

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Data insufficient

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Solution

The correct option is B $P_{1} > P_{2}$
For a one mole perfect gas, $P V = R T$ or $V = (R / P) T$
Thus, the slope of V versus T graph will be $R / P$ .
From the given diagram, slope of $P_{1}$ is less than of slope of $P_{2}$ .
So, $R / P_{1} < R / P_{2}$ or $P_{1} > P_{2}$

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The volume V versus temperature T graphs for a certain amount of a perfect gas at two pressure P1 and P2 are as shown in figure. Then

The correct option is B P1>P2For a one mole perfect gas, PV=RT or V=(R/P)TThus, the slope of V versus T graph will be R/P.From the given diagram, slope of P1 is less than of slope of P2. So, R/P1<R/P2 or P1>P2

The volume $V$ versus temperature $T$ graphs for a certain amount of a perfect gas at two pressure P $_{1}$ and P $_{2}$ are as shown in figure. Then

The correct option is B $P_{1} > P_{2}$
For a one mole perfect gas, $P V = R T$ or $V = (R / P) T$
Thus, the slope of V versus T graph will be $R / P$ .
From the given diagram, slope of $P_{1}$ is less than of slope of $P_{2}$ .
So, $R / P_{1} < R / P_{2}$ or $P_{1} > P_{2}$