The fig. given alongside shows the volume V versus temperature T graphs for a certain mass of a perfect gas at two constant pressures $P_{1}$ and $P_{2}$ . What interface can you drwa from the graphs

P_{1} > P_{2}

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P_{1} < P_{2}

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P_{1} = P_{2}

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No inference can be drawn due to insufficient information

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Solution

The correct option is A $P_{1} > P_{2}$
$P V = n R T ⟹ V = \frac{n R}{P} T$
Thus slope $= \frac{n R}{P}$
Now slope (1)< slope (2) i-e, $\frac{n R}{P_{1}} < \frac{n R}{P_{2}}$
Hence, $P_{1} > P_{2}$

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Similar questions

Q. The volume

V

versus temperature

T

graphs for a certain amount of a perfect gas at two pressures

p_{1}

and

p_{2}

are as shown in Fig. It follows from the graphs that

p_{1}

is greater than

p_{2}

Q. Volume versus temperature graphs for a given mass of an ideal gas are shown in figure at two different values of constant pressure. What can be inferred about relation between

P_{1}

and

P_{2}

A plot of volume (V) versus temperature (T) for a gas at constant pressure is a straight line passing through the origin. The plots at different values of pressure are shown in figure. Which of the following order of pressure is correct for this gas?

(a) $p_{1} > p_{2} > p_{3} > p_{4}$ (b) $p_{1} = p_{2} = p_{3} = p_{4}$

V verus T curves at constant pressures $P_{1}$ and $P_{2}$ for an idealgas are shown in the figure,

Q. In the following V-T graph, what is the relation between

P_{1}

and

P_{2}

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The fig. given alongside shows the volume V versus temperature T graphs for a certain mass of a perfect gas at two constant pressures P1 and P2. What interface can you drwa from the graphs

The correct option is A P1>P2PV=nRT⟹V=nRPTThus slope=nRPNow slope (1)< slope (2) i-e, nRP1<nRP2Hence, P1>P2

The fig. given alongside shows the volume V versus temperature T graphs for a certain mass of a perfect gas at two constant pressures $P_{1}$ and $P_{2}$ . What interface can you drwa from the graphs

The correct option is A $P_{1} > P_{2}$
$P V = n R T ⟹ V = \frac{n R}{P} T$
Thus slope $= \frac{n R}{P}$
Now slope (1)< slope (2) i-e, $\frac{n R}{P_{1}} < \frac{n R}{P_{2}}$
Hence, $P_{1} > P_{2}$