First slide

Kinetic theory of ideal gases

Question

0.014 kg of nitrogen is enclosed in a vessel at a temperature of 27 °C. Find the amount of heat energy to be supplied to the gas to double the rms speed of its molecules.

Moderate

A

4677J
B

18706J
C

9353J
D

935.3J

Solution

The root mean square speed is related to absolute temperature T as
$c_{rms} = \sqrt{\frac{3 kT}{m}}$
For a given gas, m is fixed. Therefore, $c_{rms} \propto \sqrt{T}$ . Hence in order to double the root mean square speed, the absolute temperature must be increased to four times the initial value. Initial temperature $T_{1} = 273 + 27 = 300 K$ . Therefore, final temperature $T_{2} = 4 T_{1} = 1200 K$ .
Since the volume of the vessel is fixed, $∆$ V = 0. Hence the heat energy supplied to the gas does no work on the gas; it only increases the internal energy of the molecules.
$Δ U = n C_{v} d T = \frac{m}{M} \times \frac{5 R}{2} \times (T_{2} - T_{1}) = \frac{0.014 \times 10^{3}}{28} \times \frac{5 \times 8.314}{2} \times (1200 - 300) = 9353 J$

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