Question

0.20 mol of argon gas is admitted to an evacuated 50 cm^3 container at 20 degrees Celsius. The gas then undergoes an isobaric heating to a temperature of 280 degrees Celsius.

What is the final volume of the gas in cm^3?

Answer 1

Concepts and reason

The concept used in this question is Charles law.

Firstly, find the relation between the volume and time by using the ideal gas equation.

Finally, use the expression determined above to find the new volume.

Fundamentals

The ideal gas equation is expressed as follows:

$PV = nRT$

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 of the gas.

The ideal gas equation is expressed as follows:

$PV = nRT$

Initially the system of gas is kept at pressure P, volume V₁, T₁. The ideal gas equation for this state is as follows:

$P{V_1} = nR{T_1}$

Finally, the system of gas reaches the volume V₂, temperature T₂ while the temperature remains same. The final ideal gas equation is as follows:

$P{V_2} = nR{T_2}$

Divide equation $P{V_2} = nR{T_2}$by equation $P{V_1} = nR{T_1}$and find the expression of the final volume.

$\begin{array}{c}\\\frac{{P{V_2}}}{{P{V_1}}} = \frac{{nR{T_2}}}{{nR{T_1}}}\\\\{V_2} = \frac{{{V_1}{T_2}}}{{{T_1}}}\\\end{array}$

The initial temperature in kelvin is as follows:

${T_1} = \left( {{T_1}^{\rm{o}}{\rm{ C}} + 273} \right){\rm{ K}}$

Substitute ${20^{\rm{o}}}{\rm{ C}}$ for ${T_1}$ in the above expression.

$\begin{array}{c}\\{T_1} = \left( {{{20}^{\rm{o}}}{\rm{ C}} + 273} \right){\rm{ K}}\\\\ = 2{\rm{93 K}}\\\end{array}$

The final temperature in kelvin is as follows:

${T_2} = \left( {{T_2}^{\rm{o}}{\rm{ C}} + 273} \right){\rm{ K}}$

Substitute ${280^{\rm{o}}}{\rm{ C}}$ for ${T_1}$ in the above expression.

$\begin{array}{c}\\{T_2} = \left( {{{280}^{\rm{o}}}{\rm{ C}} + 273} \right){\rm{ K}}\\\\ = 55{\rm{3 K}}\\\end{array}$

The expression of the final volume is as follows:

${V_2} = \frac{{{V_1}{T_2}}}{{{T_1}}}$

Substitute $50{\rm{ c}}{{\rm{m}}^3}$ for ${V_1}$, 553 K for ${T_2}$, and 293 K for ${T_1}$ in the above expression.

$\begin{array}{c}\\{V_2} = \frac{{\left( {50{\rm{ c}}{{\rm{m}}^3}} \right)\left( {553{\rm{ K}}} \right)}}{{\left( {293{\rm{ K}}} \right)}}\\\\ = 94.37{\rm{ c}}{{\rm{m}}^3}\\\\ = 94\,{\rm{c}}{{\rm{m}}^{\rm{3}}}\\\end{array}$

Ans:

The final volume is $94{\rm{ c}}{{\rm{m}}^3}$.