Heat and Thermodynamics 4 Question 13
15. $A 15 g$ mass of nitrogen gas is enclosed in a vessel at a temperature $27^{\circ} C$. Amount of heat transferred to the gas, so that rms velocity of molecules is doubled is about
(Take, $R=8.3 J / K$ - mol)
(2019 Main, 9 Jan II)
(a) $10 kJ$
(b) $0.9 kJ$
(c) $14 kJ$
(d) $6 kJ$
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Solution:
- We know that from kinetic theory of gases,
$$ V _{rms}=\sqrt{\frac{3 R T}{M}} \text { or } V _{rms} \propto \sqrt{T} $$
where, $R$ is gas constant, $T$ is temperature and $M$ is molecular mass of a gas.
Here, to double the $v _{rms}$, temperature must be
4 times of the initial temperature.
$\therefore$ New temperature,
$$ \begin{aligned} & T _2=4 \times T _1 \\ & \text { As, } \quad T _1=27^{\circ} C=300 K \\ & \therefore \quad T _2=4 \times 300=1200 K \end{aligned} $$
As, gas is kept inside a closed vessel.
$$ \begin{aligned} \therefore \quad Q & =n C _V \cdot d T \\ & =\frac{15}{28} \times \frac{5 R}{2}(1200-300) \\ & \quad\left[\text { As, } n=\frac{m}{M}=\frac{15}{28} \text { for nitrogen }\right] \\ & =\frac{15}{28} \times \frac{5}{2} \times 8.3 \times 900 \end{aligned} $$
[Given, $R=8.3 J / K-mole$ ]
$$ \text { or } \quad Q=10004.4 J \simeq 10 kJ $$
