Work Done During Adiabatic Expansion of Gas

To calculate the work done during the adiabatic expansion, we can use the fundamental thermodynamic relations for an adiabatic process.

1. Identify the given 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, $P_2 = 1\text{ bar} = 1 \times 10^5\text{ N/m}^2$
• Heat capacities: $c_p = 3R$ and $c_v = 2R$

2. Calculate the adiabatic index ($\gamma$): $\gamma = \frac{c_p}{c_v} = \frac{3R}{2R} = 1.5 = \frac{3}{2}$

3. Find the final volume ($V_2$): For an adiabatic process, the relationship between pressure and volume is given by: $P_1 V_1^\gamma = P_2 V_2^\gamma$

Rearranging to solve for $V_2$: $V_2 = V_1 \left( \frac{P_1}{P_2} \right)^{\frac{1}{\gamma}}$

Substitute the values: $V_2 = 0.15 \times \left( \frac{8}{1} \right)^{\frac{1}{1.5}} = 0.15 \times \left( 8 \right)^{\frac{2}{3}}$ $V_2 = 0.15 \times (2^3)^{\frac{2}{3}} = 0.15 \times 2^2 = 0.15 \times 4 = 0.6\text{ m}^3$

4. Calculate the work done ($W$) during the adiabatic process: The work done by the gas in an adiabatic expansion is given by: $W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1}$

Substituting the initial and final state values: $P_1 V_1 = (8 \times 10^5\text{ N/m}^2) \times (0.15\text{ m}^3) = 1.2 \times 10^5\text{ J}$ $P_2 V_2 = (1 \times 10^5\text{ N/m}^2) \times (0.6\text{ m}^3) = 0.6 \times 10^5\text{ J}$

Now compute $W$: $W = \frac{1.2 \times 10^5\text{ J} - 0.6 \times 10^5\text{ J}}{1.5 - 1}$ $W = \frac{0.6 \times 10^5\text{ J}}{0.5} = 1.2 \times 10^5\text{ J} = 120\text{ kJ}$

The work done during this process is 120 $\text{kJ}$.