$\frac{1}{2}$ mole of Helium gas is connected in a container at $S . T . P .$ The heat energy needed in cal to double the pressure of the gas, keeping the volume constant:

3276 J

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1638 J

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819 J

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409.5 J

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Solution

The correct option is B $1638 J$
It is given that,
No. of moles

$n = \frac{1}{2}$

$P_{1} = P$

$P_{2} = 2 P$

$T_{1} = 273 K$

$M C_{v} (m o l a r s p e c i f i c h e a t) = 3 J g^{-} K^{- 1}$

According to Gay Lussac’s law

$\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}}$

$\frac{P}{273} = \frac{2 P}{T_{2}}$

$T_{2} = 546 K$

Change in temperature

$d Q = T_{2} - T_{1}$

$= 273 K$
Now,

$d Q = n C_{v} d T$

$= \frac{1}{2} \times 12 \times 273$

$= 1638 J$

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12 mole of Helium gas is connected in a container at S.T.P. The heat energy needed in cal to double the pressure of the gas, keeping the volume constant:

The correct option is B 1638 JIt is given that,No. of moles n=12 P1=P P2=2P T1=273K MCv(molarspecificheat)=3Jg−K−1 According to Gay Lussac’s law P1T1=P2T2 P273=2PT2 T2=546K Change in temperature dQ=T2−T1 =273KNow, dQ=nCvdT =12×12×273 =1638J

$\frac{1}{2}$ mole of Helium gas is connected in a container at $S . T . P .$ The heat energy needed in cal to double the pressure of the gas, keeping the volume constant: