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

(a) Mass = 0.063 g (0° C)

(b)Mass = 0.0465 g (100° C)

(c)The pressure of the air when thermal equilibrium is reached 73 K Pa

Explanation:

Given data in the question  

$\begin{aligned}&\mathrm{V}_{1}=5 \times 10^{-5} \mathrm{m}^{3}\\&P_{1}=10^{5} \mathrm{Pa}\\&T_{1}=273 K\end{aligned}$

Molecular weight = 28.8 g

$\begin{aligned}&\mathrm{P}_{1} \mathrm{V}_{1}=\mathrm{nRT}_{1}\\&n=\frac{P_{1} V_{1}}{R T_{1}}\end{aligned}$

$\begin{aligned}&\frac{m}{M}=\frac{10^{5} \times 5 \times 10^{-5}}{8.3 \times 273}\\&\frac{m}{28.8}=\frac{10^{5} \times 5 \times 10^{-5}}{8.3 \times 273}\end{aligned}$

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

$\begin{aligned}&m=\frac{5 \times 28.8}{8.3 \times 273}\\&m=\frac{144}{2265.9}\end{aligned}$

m = 0.063 g

(b)

$\mathrm{V}_{1}=5 \times 10-5 \mathrm{m} ^3$

$\begin{aligned}&P_{1}=105 P a\\&\mathrm{P}_{2}=105 \mathrm{Pa}\end{aligned}$

$\begin{aligned}&T_{1}=273 K\\&T_{2}=373 K\end{aligned}$

M = 2 8.8 g

$\frac{P_{1} V_{1}}{T_{1}}=\frac{P_{2} V_{2}}{T_{2}}$

$\frac{5 \times 10^{-5}}{273} = \frac{V_{2}}{373}$

$\frac{5 \times 10^{-5} \times 373}{273}=V_{2}$

$V_{2}=6.831 \times 10^{-5}$

Number of air expelled = $6.831 \times 10^{-5}-5 \times 10^{-5}$

Number of air expelled  = $1.831 \times 10^{-5}$

Applying state equation, we get

PV=nRT

$\frac{m}{M}=\frac{P V}{R T}=\frac{10^{5} \times 1.831 \times 10^{-5}}{8.3 \times 373}$

$m=\frac{28.8 \times 10^{5} \times 1.831 \times 10^{-5}}{8.3 \times 373}=0.017$

Expelled air mass = 0.017 g

mass of air in the container = 0.0635 − 0.017  = 0.0465 g  

(c)  

T=273 K

P =105 Pa

$\mathrm{V}=5 \times 10^{-5} \mathrm{m}^{3}$

Applying state equation, we get

PV=nRT

$P=\frac{n R T}{V}=\frac{0.0465 \times 8.3 \times 273}{28.8 \times 5 \times 10^{-5}}$

$P=0.731 \times 10^{5} \approx 73 \mathrm{kPa}$

Therefore the mass of the air in the container of volume 50 cc at zero degree Celsius is 0.063 g and the mass is 0.0465 g at 100 degree Celsius and the pressure of the air when reached the thermal equilibrium is 73 K Pa.    

Answer 2

(a) the mass when thermal equilibrium is reached 0.0635 g

(b) the mass of air in the container when water is boiling 0.0465 g

(c) The pressure of the air when thermal equilibrium is reached 73 KPa.

Given:

Volume  = V $=50 \ \mathrm{cc}=50 \times 10^{-6} \ \mathrm{cm}^{3}$

Pressure $=100 \ \mathrm{KPa}=10^{5} \ \mathrm{Pa} \mathrm{M}=28.8 \ \mathrm{g}$

Solution:

(a) The condition for thermal equilibrium is given below.

$\mathrm{PV}=\mathrm{nRT}_{1}$

$\mathrm{PV}=\frac{\mathrm{m}}{\mathrm{M}} \mathrm{RT}_{1}$

$\Rightarrow \mathrm{m}=\frac{\mathrm{PMV}}{\mathrm{RT}_{1}}$

On substituting the values, we get,

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

$m=\frac{50 \times 28.8 \times 10^{-1}}{8.3 \times 273}$

$\therefore m=0.0635 \ \mathrm{g}$

(b) The condition when the water is boiling.

$\mathrm{PV}=\frac{\mathrm{m}}{\mathrm{M}} \mathrm{RT}_{2}$

$\mathrm{m}=\frac{\mathrm{PVM}}{\mathrm{RT}_{2}}$

On substituting the values, we get,

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

$m=\frac{50 \times 28.8 \times 10^{-1}}{8.3 \times 373}$

$\therefore m =0.0465 \ g$

(c) The container is closed and in melting-ice bath.

$\Rightarrow \mathrm{m}=\frac{\mathrm{PMV}}{\mathrm{RT}_{3}}$

On substituting the values, we get,

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

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

$\therefore P=0.07316 \times 10^{6} \ \mathrm{Pa} \approx 73 \ \mathrm{KPa}$