For a reversible adiabatic process involving ideal gas if initial pressure and volume are 8 bar and 0.15 m³ respectively and final pressure is 1 bar. Calculate |work done| (in Kilo Joule) [C_V = 2_R, C_P = 3R]

The correct answer is 120.

Let us analyze the reversible adiabatic process step-by-step.

Step 1: Determine the adiabatic index ($\gamma$)

The heat capacity at constant volume is given as:

$C_V=2R$

The heat capacity at constant pressure is given as:

$C_P=3R$

The adiabatic index ($\gamma$) is the ratio of specific heat capacities:

$\gamma =\frac{C_P}{C_V}$

Substituting the values:

$\mathrm{��}=\frac{3R}{2R}$

Step 2: Find the final volume ($V_2$)

For a reversible adiabatic process involving an ideal gas, the pressure and volume are related by:

$P_1{V}_1^{\gamma }=P_2{V}_2^{\gamma }$

Rearranging the equation to solve for the final volume $V_2$ yields:

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

Given the initial and final states:

Initial pressure, $P_1=8\text{bar}$

Initial volume, $V_1=0.15{\text{m}}^3$

Final pressure, $P_2=1\text{bar}$

Substituting these values into the equation:

$V_2=0.15\times {\left(\frac{8}{1}\right)}^{\frac{1}{1.5}}$

Since $\frac{1}{1.5}=\frac{2}{3}$:

$V_2=0.15\times 8^{\frac{2}{3}}$

$V_2=0.15\times {\left(8^{\frac{1}{3}}\right)}^2$

$V_2=0.15\times 2^2$

$V_2=0.15\times 4=0.60{\text{m}}^3$

Step 3: Calculate the work done ($W$)

The work done during a reversible adiabatic process is given by the formula:

$W=\frac{P_2{V}_2-P_1{V}_1}{\gamma -1}$

Converting the pressures from bar to Pascals ($1\text{bar}={10}^5{\text{N/m}}^2$):

$P_1=8\times {10}^5{\text{N/m}}^2$

$P_2=1\times {10}^5{\text{N/m}}^2$

Substituting the values into the work formula:

$W=\frac{(1\times {10}^5\times 0.60)-(8\times {10}^5\times 0.15)}{1.5-1}$

$W=\frac{60000-120000}{0.5}$

$W=\frac{-60000}{0.5}=-120000\text{J}$

Converting the work done to Kilo Joules (kJ):

$W=-120\text{kJ}$

Taking the absolute value of the work done:

$\left|W\right|=120\text{kJ}$