One mole of a gas at a pressure 2 Pa and temperature 27 $^{o}$ C is heated till both pressure and volume are doubled. What is the temperature of the gas?

1200 K

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900 K

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600 K

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300 K

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Solution

The correct option is B 1200 K
We have the relation PV=nRT (ideal gas equation)

T= $\frac{P V}{n R}$

Given, that pressure and volume are doubled i.e, $P_{1} = 2 P$ , $V_{1} = 2 V$ .

$T_{1} = \frac{P_{1} V_{1}}{n R}$

$= \frac{(2 P \times 2 V)}{n R}$

$= 4 T$

Given that T=27 $^{o}$ C i.e, 300K

$T_{1} = 4 \times 300$

$= 1200 K$

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One mole of a gas at a pressure 2 Pa and temperature 27oC is heated till both pressure and volume are doubled. What is the temperature of the gas?

The correct option is B 1200 KWe have the relation PV=nRT (ideal gas equation)T=PVnRGiven, that pressure and volume are doubled i.e, P1=2P,V1=2V.T1=P1V1nR=(2P×2V)nR=4TGiven that T=27oC i.e, 300KT1=4×300=1200K

One mole of a gas at a pressure 2 Pa and temperature 27 $^{o}$ C is heated till both pressure and volume are doubled. What is the temperature of the gas?