Question

A 3.00-L tank contains air at 3.00atm and 20.0 C. The tank is sealed and cooled until the pressure is 1.00atm.

(A)What is the temperature then in degrees Celsius? Assume that the volume of the tank is constant.

(B)If the temperature is kept at the value found in part A and the gas is compressed, what is the volume when the pressure again becomes 3.00atm?

Answer 1

Concepts and reason

The concepts used to solve the given problem are ideal gas equation and Boyle’s law.

Initially, derive the expression for the final temperature by using Boyle’s law. Next, Calculate the final temperature by using the relation between pressure and temperature. After that, derive the expression for the final volume by using the ideal gas equation. Finally, calculate the final volume by using the relation between pressure and volume.

Fundamentals

The ideal gas equation is,

$PV = nRT$

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

According to Boyle’s law, at the constant temperature, the volume of a mass is inversely proportional to the pressure.

$V \propto \frac{1}{P}$

The relation between pressure and temperature is,

$\frac{{{P_1}}}{{{P_2}}} = \frac{{{T_1}}}{{{T_2}}}$

Here, ${P_1}$ is the initial pressure, ${P_2}$ is the final pressure, ${T_1}$ is initial temperature, and ${T_2}$ is the final temperature.

The relation between pressure and volume is,

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

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

The final temperature.

${T_2} = \left( {\frac{{{P_2}}}{{{P_1}}}} \right){T_1}$

Substitute 1 atm for ${P_2}$ , 3 atm for ${P_1}$ , and 293.15 K for ${T_1}$ .

$\begin{array}{c}\\{T_2} = \left( {\frac{{1\;{\rm{atm}}}}{{3\;{\rm{atm}}}}} \right)\left( {293.15\;{\rm{K}}} \right)\\\\ = 97.72\;{\rm{K}} - 273.15\;{\rm{K}}\\\\{\rm{ = }} - {\rm{175}}{\rm{.43}}\;^\circ {\rm{C}}\\\end{array}$

(B)

The final volume of the gas is,

${V_2} = \left( {\frac{{{P_1}}}{{{P_2}}}} \right){V_1}$

Substitute 1 atm for ${P_2}$ , 3 atm for ${P_1}$ , and 3 L for ${V_1}$ .

$\begin{array}{c}\\{V_2} = \left( {\frac{{1\;{\rm{atm}}}}{{3\;{\rm{atm}}}}} \right)\left( {3\;{\rm{L}}} \right)\\\\ = 1\;{\rm{L}}\\\end{array}$

Ans: Part A

The final temperature is $- {\rm{175}}{\rm{.43}}\;^\circ {\rm{C}}$ .

Part B

The final volume of the gas is 1 L.