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°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°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

$\begin{array}{l}\left(\text{a}\right)\text{Here,}\\ V_1=5\times {10}^{-5}{m}^3\\ P_1={10}^5\text{Pa}\\ T_1=273\text{K}\\ \text{M = 28}\text{.8 g}\\ P_1{V}_1=nR{T}_1\\ \Rightarrow n=\frac{P_1{V}_1}{R{T}_1}\\ \Rightarrow \frac{m}{M}=\frac{{10}^5\times 5\times {10}^{-5}}{8.3\times 273}\\ \Rightarrow m=\frac{{10}^5\times 5\times {10}^{-5}\times 28.8}{8.3\times 273}\\ \Rightarrow m=0.0635\text{g}\end{array}$

$\begin{array}{l}\left(\text{b}\right)\mathrm{Here},\\ V_1=5\times {10}^{-5}{m}^3\\ P_1={10}^5\text{Pa}\\ P_2={10}^5\text{Pa}\\ T_1=273\text{K}\\ T_2=373\text{K}\\ \text{M = 28}\text{.8 g}\\ \frac{P_1{V}_1}{T_1}=\frac{P_2{V}_2}{T_2}\\ \Rightarrow \frac{5\times {10}^{-5}}{273}=\frac{V_2}{373}\\ \Rightarrow V_2=\frac{5\times {10}^{-5}\times 373}{273}\\ \Rightarrow V_2=6.831\times {10}^{-5}\\ \text{Volume of expelled air =}6.831\times {10}^{-5}-5\times {10}^{-5}\\ =1.831\times {10}^{-5}\\ \text{Applying equation of state, we get}\\ PV=nRT\\ \Rightarrow \frac{m}{M}=\frac{PV}{RT}=\frac{{10}^5\times 1.831\times {10}^{-5}}{8.3\times 373}\\ \Rightarrow m=\frac{28.8\times {10}^5\times 1.831\times {10}^{-5}}{8.3\times 373}=0.017\\ \text{Thus, mass of expelled air = 0}\text{.017 g}\\ \text{Amount of air in the container = 0}\text{.0635}-\text{0}\text{.017}\\ =0.0465\text{g}\end{array}$

$\begin{array}{l}\left(\text{c}\right)\text{Here,}\\ T=273K\\ P={10}^5\text{Pa}\\ \text{V=5}\times {\text{10}}^{-5}{\text{m}}^3\\ \text{Applying equation of state, we get}\\ PV=nRT\\ \Rightarrow P=\frac{nRT}{V}=\frac{0.0465\times 8.3\times 273}{28.8\times 5\times {10}^{-5}}\\ \Rightarrow P=0.731\times {10}^5\approx 73\text{kPa}\end{array}$