A container of initial volume 0.15m³ is expanded adiabatically from pressure 8 bar to final pressure 1 bar. If initial temperature is 140K, find work done by the gas. (C_p = 3R, C_v = 2R)

The correct answer is 120kJ.

Step 1: Identify the given values and parameters
Initial volume, $V_1=0.15{\text{m}}^3$
Initial pressure, $P_1=8\text{bar}=8\times {10}^5{\text{N/m}}^2$
Final pressure, $P_2=1\text{bar}=1\times {10}^5{\text{N/m}}^2$
Initial temperature, $T_1=140\text{K}$
Specific heat capacity at constant pressure, $C_p=3R$
Specific heat capacity at constant volume, $C_v=2R$

Step 2: Calculate the adiabatic index ($\gamma$)
The ratio of specific heats ($\gamma$) is given by:
$\gamma =\frac{C_p}{C_v}=\frac{3R}{2R}=1.5$

Step 3: Determine the final volume ($V_2$)
For an adiabatic process, the relationship between pressure and volume is:
$P_1{V}_1^{\gamma }=P_2{V}_2^{\gamma }$
Rearranging this formula to find the ratio of volumes:
${\left(\frac{V_2}{V_1}\right)}^{\gamma }=\frac{P_1}{P_2}$
Substitute the values of pressure and $\gamma$ into the equation:
${\left(\frac{V_2}{V_1}\right)}^{1.5}=\frac{8}{1}=8$
Since $1.5=\frac{3}{2}$, we can solve for $\frac{V_2}{V_1}$ by raising both sides to the power of $\frac{2}{3}$:
$\frac{V_2}{V_1}=8^{2/3}={\left(2^3\right)}^{2/3}=2^2=4$
Therefore, the final volume $V_2$ is:
$V_2=4\times V_1=4\times 0.15{\text{m}}^3=0.6{\text{m}}^3$

Step 4: Calculate the work done by the gas ($W$)
The work done during an adiabatic process is given by the formula:
$W=\frac{P_1{V}_1-P_2{V}_2}{\gamma -1}$
Let us compute the individual terms in the numerator:
$P_1{V}_1=\left(8\times {10}^5{\text{N/m}}^2\right)\times \left(0.15{\text{m}}^3\right)=1.2\times {10}^5\text{J}$
$P_2{V}_2=\left(1\times {10}^5{\text{N/m}}^2\right)\times \left(0.6{\text{m}}^3\right)=0.6\times {10}^5\text{J}$
Now, substitute these values back into the work done equation:
$W=\frac{1.2\times {10}^5-0.6\times {10}^5}{1.5-1}$
$W=\frac{0.6\times {10}^5}{0.5}=1.2\times {10}^5\text{J}$
Converting the work done to kilojoules (kJ):
$W=120\text{kJ}$