Question

The air temperature and pressure in a laboratory are 20$°$C and 1.0 atm. A 1.0 L container is open to the air. The container is then sealed and placed in a bath of boiling water. After reaching thermal equilibrium, the container is opened. How many moles of air escape?

Short answer

$\Delta \mathrm{n}=0.0113\mathrm{mol}$is the number of moles of air that escaped from the container.

Step-by-step solution

Given information

The temperature is$T=20+273=293K$$T=20+273=293K$

Pressure $P=1\mathrm{atm}$

The thermal temperature of equilibrium $T_2=100+273=373K$

It's vital to remember that the number of moles doesn't change while they're immersed in boiling water. When the container is opened, the pressure equals $1\mathrm{atm}$

formula used:

Ideal gas law

$PV=nRT$

Calculation

Description	Steps
Step:1 The number moles calculating	$$\begin{gathered} n=\frac{PV}{RT} \\ =\frac{\text{latm}\left(\mathrm{LL}\right)}{293K\left(0.082057{\mathrm{LatmK}}^{-1}{\mathrm{mol}}^{-1}\right)} \\ n=0.0415\mathrm{mol} \\ \text{P, V, R are constants} \\ \frac{n_2}{n_1}=\frac{T_1}{T_2} \\ {n}_2=\frac{273}{373}(0.0415) \\ {n}_2=0.0528\mathrm{mol} \end{gathered}$$
Step:2 Counting the number of moles who have gotten away	$$\begin{gathered} \Delta n=n_2-n_1 \\ =0.0528\mathrm{mol}-0.0415\mathrm{mol} \\ \Delta n=0.0113\mathrm{mol} \end{gathered}$$ $$\begin{gathered} \Delta n=n_2-n_1 \\ =0.0528mol-0.0415mol \\ \Delta n=0.0113mol \end{gathered}$$

$\Delta n=0.0113$is the number of moles of air that escaped from the container.