$1 / 2$ mole of helium is contained in a container at STP. How much heat energy is needed to double the pressure of the gas, (volume is constant) heat capacity of gas is $3$ J $g^{- 1} K^{- 1}$ .

1436

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736

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1638

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5698

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Solution

The correct option is B $1638$ J
Here, $n = \frac{1}{2}, c_{v} = 3 J g^{- 1} K^{- 1}$ , M $= 4$ g $m o l^{- 1}$
$∴ C_{v} = M_{C_{V}} 4 \times 3 = 12$ J $m o l^{- 1}$ $K^{- 1}$
At constant volume $P \propto T$ .
$∴ \frac{P_{2}}{P_{1}} = \frac{T_{2}}{T_{1}} = 2, T_{2} = 2 T_{1}$
Rise in temperature $Δ T = T_{2} - T_{1} = 2 T_{1} - T_{1} = T_{1} = 273$ K
Heat required, $Δ Q = N C_{V} Δ T = \frac{1}{2} \times 12 \times 273 = 1638$ J

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1/2 mole of helium is contained in a container at STP. How much heat energy is needed to double the pressure of the gas, (volume is constant) heat capacity of gas is 3J g−1K−1.

The correct option is B 1638JHere, n=12,cv=3Jg−1K−1, M=4g mol−1∴Cv=MCV4×3=12J mol−1 K−1At constant volume P∝T.∴P2P1=T2T1=2,T2=2T1Rise in temperature ΔT=T2−T1=2T1−T1=T1=273KHeat required, ΔQ=NCVΔT=12×12×273=1638J

$1 / 2$ mole of helium is contained in a container at STP. How much heat energy is needed to double the pressure of the gas, (volume is constant) heat capacity of gas is $3$ J $g^{- 1} K^{- 1}$ .