Helium is contained in a closed vessel with a volume of $2$ litres at a temperature of $20^{\circ} C$ and a pressure of $10^{5} N / m^{2}$ .
(a) What amount of heat should be imparted to the helium to increase its temperature to $100^{\circ} C$ ?
(b) What will be the r.m.s. velocity of its molecules be at the new temperature?
(c) What will the pressure be?
(d) What is the density of the helium?

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Solution

Initial pressure $= 10^{5} N / m^{2}$
Initial volume $= 2 l t = 2 \times 1000 c . c . = 0.002 m^{3}$
Initial temperature $= 293 K$
As the vessel is closed, the process is a constant volume process.
Final temperature $= 100^{\circ} C = 373 K$
Number of moles $= \frac{10^{5} \times 0.02}{8.314 \times 293} = 0.0821 m o l e s$
Specific heat at constant volume, $C_{V} = \frac{3 R}{2} = 12.4 J / m o l e / K$
(a) Hence, heat to be added $= (0.0821) (12.47) (80) = 81.91 J$
(b) Molecular weight of Helium : $M = 4 \times 10^{- 3} k g$
$V_{r m s} = \sqrt{\frac{3 R T}{M}} = 1525 m / s$
(c) $\frac{P_{1}}{T_{1}} = \frac{P_{2}}{T_{2}} \Rightarrow \frac{P_{2}}{373} = \frac{10^{5}}{293} N / m^{2}$
$\Rightarrow P_{2} = 1.27 \times 10^{5} N / m^{2}$
(d) $P V = \frac{m}{M} R T$
$\Rightarrow \frac{P M}{R T} = ρ = 0.164 k g / m^{3}$ .

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