Question

Calculate the work done when one mole of a perfect gas is compressed adiabatically. The initial pressure and volume of the gas are ${10^5}N/{m^2}$ and 6L respectively. The final volume of the gas is 2L. the molar specific heat of the gas at constant volume is $\dfrac{{3R}}{2}$

Answer 1

An adiabatic process is a process in which heat is not allowed to leave or enter the system (i.e. no heat exchange with the surroundings). In such a process, both pressures of the system will change with the volume.
Formula used: In this solution we will be using the following formulae;
$P{V^\gamma } = constant$ where $P$ stands for pressure and $V$ for volume, $\gamma$ is the adiabatic constant.
$\gamma = \dfrac{{{c_p}}}{{{c_v}}}$ where ${c_p}$ is the specific heat capacity of a gas at constant pressure, and ${c_v}$ is the specific heat capacity at constant volume.
${c_p} - {c_v} = R$ where $R$ is the molar gas constant.
$W = \dfrac{{{P_2}{V_2} - {P_1}{V_1}}}{{1 - \gamma }}$ where $W$ is the work done by a gas in an adiabatic process, the subscript 2 and 1 signifies the final and initial state of the system.

Complete Step-by-Step solution:
For adiabatic process, we have that
$P{V^\gamma } = constant$ where $P$ stands for pressure and $V$ for volume, $\gamma$ is the adiabatic constant
Hence, by comparison on one state to another, we may have
${P_1}{V_2}^\gamma = {P_2}{V_2}^\gamma$
But $\gamma = \dfrac{{{c_p}}}{{{c_v}}}$ where ${c_p}$ is the specific heat capacity of a gas at constant pressure, and ${c_v}$ is the specific heat capacity at constant volume.
and again, ${c_p} - {c_v} = R$ where $R$ is the molar gas constant.
Hence, by inserting values
${c_p} - \left( {\dfrac{{3R}}{2}} \right) = R$
$\Rightarrow {c_p} = R + \dfrac{{3R}}{2} = \dfrac{{5R}}{2}$
Hence, the adiabatic constant can be calculated as
$\gamma = \dfrac{{{c_p}}}{{{c_v}}} = \dfrac{{5R}}{2} \div \dfrac{{3R}}{2}$
$\Rightarrow \gamma = \dfrac{5}{3}$
Hence inserting into ${P_1}{V_2}^\gamma = {P_2}{V_2}^\gamma$, we have
$\left( {{{10}^5}} \right){\left( 6 \right)^{\dfrac{5}{3}}} = {P_2}{\left( 2 \right)^{\dfrac{5}{3}}}$
Hence, by dividing both sides by ${\left( 2 \right)^{\dfrac{5}{3}}}$ we have
${P_2} = \left( {{{10}^5}} \right){\left( 3 \right)^{\dfrac{5}{3}}} = 6.19 \times {10^5}N/{m^2}$
The work done in an adiabatic process is given by
$W = \dfrac{{{P_2}{V_2} - {P_1}{V_1}}}{{1 - \gamma }}$
Hence, inserting all known values, we get
$W = \dfrac{{6.19 \times {{10}^5}\left( {2 \times {{10}^{ - 3}}} \right) - {{10}^5}\left( {6 \times {{10}^{ - 3}}} \right)}}{{1 - \dfrac{5}{3}}}$ (since 1000 Litre is 1 cubic metre).
Computing the equation, we have
$W = - 957J$
Negative signifies work is done on the system.

Note: For clarity, observe that in the relation ${P_1}{V_2}^\gamma = {P_2}{V_2}^\gamma$ we do not have to convert to SI units. This is because it leads to a ratio of the volumes and is hence units along with any conversion factor will cancel out eventually.