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The amount of heat energy required to raise the temperature of 1 g of Helium at NTP, from $T_{1} K t o T_{2} K$ is

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$\frac{3}{4} N_{a} k_{B} (T_{2} - T_{1})$

$\frac{3}{4} N_{a} k_{B} (\frac{T_{2}}{T_{1}})$

$\frac{3}{8} N_{a} k_{B} (T_{2} - T_{1})$

$\frac{3}{2} N_{a} k_{B} (T_{2} - T_{1})$

answer is C.

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Detailed Solution

As here volume of the gas remains constant, therefore the amount of heat energy required to raise the temperature of the gas is $Δ Q = n C_{v} Δ T$

Here, Number of moles, $n = \frac{1}{4}$

$C_{V} = \frac{3}{2} R (∵ H e i s a m o n a t o m i c) Δ T = T_{2} - T_{1} ∴ Δ Q = \frac{1}{4} \frac{3}{2} R (T_{2} - T_{1}) = \frac{3}{8} N_{a} k_{B} (T_{2} - T_{1}) (∵ k_{B} = \frac{R}{N_{a}})$

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