A cylinder with fixed capacity of 67.2 , L co | vedclass

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(C) At $STP$,the molar volume of an ideal gas is $22.4 \, L/mol$.The number of moles $n$ of helium gas is $n = \frac{67.2 \, L}{22.4 \, L/mol} = 3 \,

Vedclass · Accepted Answer

$748$

Vedclass · Answer

(C) At $STP$,the molar volume of an ideal gas is $22.4 \, L/mol$.The number of moles $n$ of helium gas is $n = \frac{67.2 \, L}{22.4 \, L/mol} = 3 \, mol$.Helium is a monoatomic gas,so its molar heat capacity at constant volume is $C_v = \frac{3}{2}R$.The heat required is given by $\Delta Q = n C_v \Delta T$.Substituting the values: $\Delta Q = 3 	imes \left(\frac{3}{2} 	imes 8.31 ight) 	imes 20$.$\Delta Q = 3 	imes 1.5 	imes 8.31 	imes 20 = 4.5 	imes 166.2 = 747.9 \, J$.Rounding to the nearest integer,we get $\Delta Q \approx 748 \, J$.

Molecule type	$\frac{C_{P}}{C_{v}}$
$A$. Monoatomic	$I$. $\frac{7}{5}$
$B$. Diatomic rigid molecules	$II$. $\frac{9}{7}$
$C$. Diatomic non-rigid molecules	$III$. $\frac{4}{3}$
$D$. Triatomic rigid molecules	$IV$. $\frac{5}{3}$

$A$ cylinder with fixed capacity of $67.2 \, L$ contains helium gas at $STP$. The amount of heat needed to raise the temperature of the gas by $20 \, ^oC$ is ..... $J$ [Given that $R = 8.31 \, J \, mol^{-1} \, K^{-1}$]

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