Question

A rigid vessel of the volume $0.50m^3$ containing H₂​ at $20.50{}^{o}C$ and pressure of $611\times {10}^{3}Pa$ was connected to a second rigid vessel of the volume$0.75m^3$ containing Ar at$31.20{}^{o}C$ at a pressure of $433\times {10}^{3}Pa$. A valve separating the two vessels is opened and both are cooled to a temperature of $14.50{}^{o}C$. What is the final pressure in the vessels?

Answer 1

Given data

1. The volume of the first and second vessels are $v_1=0.50m^3$ and $v_2=0.75m^3$.20.50
2. The temperature of the gases in the first and second vessels are $T_1=20.{50}^{o}C$ and $T_2=31.20{}^{o}C$.
3. The pressure on the first and second vessels are $p_1=611\times {10}^{3}pa$ and $p_2=433\times {10}^{3}pa.$

Boyle's law for ideal gas

1. Boyle's law states that the pressure of a gas is inversely proportional to its volume at a constant temperature.
2. Boyle's law is defined by the form, $pv=nRT$, where, p and v are the pressure and volume of the gas, T is the temperature, n is the number of moles and R is the universal gas constant.

Finding the pressure of the gas mixture

We know, for an ideal gas,

$$\begin{gathered} pv=nRT \\ orn=\frac{pv}{RT} \end{gathered}$$

It is clear that the number of gas molecules is the same before and after the mixture. Let p, T, and v be the pressure, temperature and volume of the mixture gas.

So,

$$\begin{gathered} \frac{p_1{v}_1}{R{T}_1}+\frac{p_2{v}_2}{R{T}_2}=\frac{pv}{RT} \\ or\frac{\left(611\times {10}^3\right)\times 0.50}{\left(20.50+273\right)}+\frac{\left(433\times {10}^3\right)\times 0.75}{1067.56\left(31.20+273\right)}=\frac{p\times \left(0.5+0.75\right)}{\left(14.50+273\right)}\left(\sin ce,20.50{}^{o}C=\left(20.50+273\right)K\right) \\ or1040.89+1067.56=4.35\times {10}^{-3}p\left(Andv=(v_1+v_2)\right) \\ orp=\frac{2108.45}{4.35\times {10}^{-3}}=4.84\times {10}^{5}pa. \\ orp=4.84\times {10}^{5}pa. \end{gathered}$$

So, the final pressure in the vessels is $4.84\times {10}^{5}pa.$