Question

A container of volume 50 cc contains air (mean molecular weight = 28.8 g) and is open to atmosphere where the pressure is 100 kPa. The container is kept in a bath containing melting ice ($0^0$C). (a) Find the mass of the air in the container when thermal equilibrium is reached. (b) The container is now placed in another bath containing boiling water (${100}^0$C). Find the mass of air in the container. (c) The container is now closed and placed in the melting-ice bath. Find the pressure of the air when thermal equilibrium is reached.

Answer 1

Given , V = 50 cc. = $50\times {10}^{-6}{m}^{-3}$

P = 100 KPa = ${10}^5$ Pa,

M= 28.8 g

(a) PV = $nR{T}_1$

$\Rightarrow PV=\frac{m}{M}R{T}_1$

$\Rightarrow m=\frac{PMV}{R{T}_1}$

= $\frac{{10}^5\times 28.8\times 50\times {10}^{-6}}{8.3\times 273}$

= 0.0635 g

(b) When the vessel is kept on boiling water

$PV=\frac{nR{T}_2}{M}$

$\Rightarrow m=\frac{PVM}{R{T}_2}$

= $\frac{{10}^5\times 28.8\times 50\times {10}^{-6}}{8.3\times 373}$

= 0.0465 gm.

(c) When the vessel is closed

$P\times 50\times {10}^{-6}=\frac{0.0465}{28.8}\times 8.3\times 273$

$\Rightarrow P=\frac{0.0465\times 8.3\times 273}{28.8\times 50\times {10}^{-6}}$

= $0.07316\times {10}^6$ Pa

= 73.16 KPa

= 73 KPa.