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The amount of heat energy required to raise the temperature of $1\, g$ of helium from $T_1\,K$ to $T_2K$ is
$\frac{3}{2}\,{N_a}{k_B}\,\left( {{T_2} - {T_1}} \right)$
$\frac{3}{4}\,{N_a}{k_B}\,\left( {{T_2} - {T_1}} \right)$
$\frac{3}{4}\,{N_a}{k_B}\,\left( {\frac{{{T_2}}}{{{T_1}}}} \right)$
$\frac{3}{8}\,{N_a}{k_B}\,\left( {{T_2} - {T_1}} \right)$
Solution
As here volume of the gas remains constant, therefore, the amount of heat energy to raise the temperature of the gas is :
$\Delta Q=n C_{v} \Delta T$
Here, number of moles, $n=\frac{1}{4}$
$C_{v}=\frac{3}{2} R \quad(\because \text { He is a monoatomic gas })$
$\Delta T=T_{2}-T_{1}$
$\therefore \Delta Q=\frac{1}{4}\left(\frac{3}{2} R\right)\left(T_{2}-T_{1}\right)$
$=\frac{3}{8} N_{a} K_{B}\left(T_{2}-T_{1}\right) \quad\left[\because k_{B}=\frac{R}{N_{a}}\right]$
Hence, correct answer is $( 4)$
