Qstnid -Iv-32-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. - 24: Kinetic Theory Of Gases

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NEET-XII-Physics

24: Kinetic Theory of Gases

with Solutions - page 6

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#32
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.
Ans : `` \begin{array}{l}\left(\text{a}\right)\text{}\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}``
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`` \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{\text{}5\times {10}^{-5}}{273}=\frac{{V}_{2}}{373}\\ \Rightarrow {V}_{2}=\frac{\text{}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}``
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