Question

The volume of a gas is halved during an adiabatic compression that increases the pressure by a factor of 2.5. What is the specific heat ratio?

Answer 1

Concepts and reason

The concept used to solve this problem is the ideal gas equation in a reversible adiabatic process.

Use the ideal gas equation to calculate the specific heat ratio.

Fundamentals

Expression for the ideal gas equation in a reversible adiabatic process is,

$P{V^n} = {\rm{constant}}$

But here $n = \gamma$ , since the equation becomes,

$P{V^\gamma } = {\rm{constant}}$

Here, pressure is $P$ , volume is $V$ , and the specific heat ratio is $\gamma$ .

Expression for the relation between the initial pressure and volume to the final pressure and volume is,

${P_i}V_i^\gamma = {P_f}V_f^\gamma$

Here, initial pressure is ${P_i}$ , initial volume is ${V_i}$ , final pressure is ${P_f}$ , final volume is ${V_f}$ and the specific heat ratio is $\gamma$ .

The pressure is increased by a factor and volume is halved during the adiabatic process.

Expression for the relation between initial pressure and volume to final pressure and volume is,

${P_i}V_i^\gamma = {P_f}V_f^\gamma$

The product of pressure and volume in an adiabatic process is constant.

Which means that initial pressure multiplied with initial volume is equal to final pressure multiplied with volume.

Expression for relation between initial pressure and volume to final pressure and volume is,

${P_i}V_i^\gamma = {P_f}V_f^\gamma$

Substitute $2.5{P_i}$ for ${P_f}$ and $0.5{V_i}$ for ${V_f}$ .

${P_i}V_i^\gamma = \left( {2.5{P_i}} \right){\left( {0.5{V_i}} \right)^\gamma }$

Rearrange the above relation.

$\begin{array}{c}\\\frac{{{P_i}}}{{2.5{P_i}}} = {\frac{{\left( {0.5{V_i}} \right)}}{{V_i^y}}^\gamma }\\\\\frac{{{P_i}}}{{2.5{P_i}}} = {\left( {\frac{{0.5{V_i}}}{{{V_i}}}} \right)^\gamma }\\\\\frac{1}{{2.5}} = {\left( {0.5} \right)^\gamma }\\\\0.4 = {\left( {0.5} \right)^\gamma }\\\end{array}$

Take natural logarithm on both sides,

$\begin{array}{c}\\\ln \left( {0.4} \right) = \ln {\left( {0.5} \right)^\gamma }\\\\\gamma \ln \left( {0.5} \right) = \ln \left( {0.4} \right)\\\end{array}$

The specific heat ratio is,

$\begin{array}{c}\\\gamma = \frac{{\ln \left( {0.4} \right)}}{{\ln \left( {0.5} \right)}}\\\\ = \frac{{ - 0.916}}{{ - 0.693}}\\\\ = 1.322\\\end{array}$

Therefore, the specific heat ratio is $\gamma = 1.322$ .

Ans:

Thus, the specific heat ratio of the gas is ${\bf{\gamma = 1}}{\bf{.322}}$ .