Question

0.16 mol of argon gas is admitted to an evacuated 70 cm^3 container at 20 C . The gas then undergoes an isothermal expansion to a volume of 500 cm^3. What is the final pressure of the gas in atm?

Answer 1

Concepts and reason

The concept required to solve this problem is the ideal gas equation.

Initially, write the expression of the ideal gas equation. Using the number of moles, volume, and temperature calculate the initial pressure. Later, after thermal expansion the final pressure can be calculated using the initial temperature and the volume after the expansion.

Fundamentals

The expression of the ideal gas is,

$PV = nRT$

Here, P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature.

The ideal gas equation is expressed as follows:

${P_1}{V_1} = {P_2}{V_2}$

Here, ${P_1}$is the initial pressure, ${V_1}$ is the initial volume, ${P_2}$ is the final pressure, and ${V_2}$ is the final volume.

The expression of the ideal gas is,

${P_1}{V_1} = nRT$

Here, ${P_1}$ is the initial pressure, and ${V_1}$ is the initial volume.

Rearrange the expression for ${P_1}$.

${P_1} = \frac{{nRT}}{{{V_1}}}$

Substitute 0.16 mol for n, $8.314{\rm{ J/mol}} \cdot {\rm{K}}$ for R, $20^\circ {\rm{ C}}$ for T, and $70{\rm{ c}}{{\rm{m}}^3}$ for ${V_1}$.

$\begin{array}{c}\\{P_1} = \frac{{\left( {0.16{\rm{ mol}}} \right)\left( {8.314{\rm{ J/mol}} \cdot {\rm{K}}} \right)\left( {\left( {20 + 273} \right){\rm{ K}}} \right)}}{{70{{\left( {{\rm{cm}}\left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1{\rm{ cm}}}}} \right)} \right)}^3}}}\\\\ = \left( {5.568 \times {{10}^6}{\rm{ Pa}}} \right)\left( {\frac{{9.86 \times {{10}^{ - 6}}{\rm{ atm}}}}{{1{\rm{ Pa}}}}} \right)\\\\ = 54.9{\rm{ atm}}\\\end{array}$

The temperature remains constant in the thermal expansion.

Use Ideal gas equation.

${P_1}{V_1} = {P_2}{V_2}$

Here, ${P_1}$is the initial pressure, ${V_1}$ is the initial volume, ${P_2}$ is the final pressure, and ${V_2}$ is the final volume.

Rearrange the equation for ${P_2}$.

${P_2} = \frac{{{P_1}{V_1}}}{{{V_2}}}$

Substitute $54.9{\rm{ atm}}$ for ${P_1}$, $70{\rm{ c}}{{\rm{m}}^3}$ for ${V_1}$, and $500{\rm{ c}}{{\rm{m}}^3}$ for ${V_2}$.

$\begin{array}{c}\\{P_2} = \frac{{\left( {54.9{\rm{ atm}}} \right)\left( {70{{\left( {{\rm{cm}}\left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1{\rm{ cm}}}}} \right)} \right)}^3}} \right)}}{{500{{\left( {{\rm{cm}}\left( {\frac{{{{10}^{ - 2}}{\rm{ m}}}}{{1{\rm{ cm}}}}} \right)} \right)}^3}}}\\\\ = 7.69{\rm{ atm}}\\\end{array}$

Ans:

The final pressure of the gas is 7.69 atm.