A gas at $27^{\circ} C$ has a volume V and pressure P. On heating its pressure is doubled and volume becomes three times. The resulting temperature of the gas will be

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Solution

The correct option is C
From ideal gas equation $P V = μ R T w e g e t \frac{T_{2}}{T_{1}} = (\frac{P_{2}}{P 1}) (\frac{V_{2}}{V_{1}}) = (\frac{2 P_{1}}{P_{1}}) (\frac{3 V_{1}}{V_{1}}) = 6$
$∴ T_{2} = 6 T_{1} = 6 \times 300 = 1800 K = 1527^{\circ} C$

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A gas at 27∘C has a volume V and pressure P. On heating its pressure is doubled and volume becomes three times. The resulting temperature of the gas will be

The correct option is C From ideal gas equation PV=μRT we getT2T1=(P2P1)(V2V1)=(2P1P1)(3V1V1)=6 ∴T2=6T1=6×300=1800 K=1527∘C

A gas at $27^{\circ} C$ has a volume V and pressure P. On heating its pressure is doubled and volume becomes three times. The resulting temperature of the gas will be

The correct option is C
From ideal gas equation $P V = μ R T w e g e t \frac{T_{2}}{T_{1}} = (\frac{P_{2}}{P 1}) (\frac{V_{2}}{V_{1}}) = (\frac{2 P_{1}}{P_{1}}) (\frac{3 V_{1}}{V_{1}}) = 6$
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