As shown in Fig. P2.59, a gas contained within a piston-cylinder assembly, initially at a volume of 0.1 m³, undergoes a constant-pressure expansion at 2 bar to a final volume of 0.... As shown in Fig. P2.59, a gas contained within a piston-cylinder assembly, initially at a volume of 0.1 m³, undergoes a constant-pressure expansion at 2 bar to a final volume of 0.12 m³, while being slowly heated through the base. The change in internal energy of the gas is 0.25 kJ. The local atmospheric pressure is 1 bar. (a) For the gas as the system, evaluate work and heat transfer, each in kJ. (b) For the piston as the system, evaluate work and change in potential energy, each in kJ. Read more

Steps to Solve

1. Calculate Work for the Gas as the System

The work done by the gas during expansion at constant pressure can be calculated using the formula: $$ W = P \cdot \Delta V $$ where:

• $P = 2 \text{ bar} = 200 \text{ kPa}$ (since 1 bar = 100 kPa)
• $\Delta V = V_2 - V_1 = 0.12 , \text{m}^3 - 0.1 , \text{m}^3 = 0.02 , \text{m}^3$

Substituting the values: $$ W = 200 , \text{kPa} \times 0.02 , \text{m}^3 = 4 , \text{kJ} $$

2. Determine Heat Transfer for the Gas as the System

Using the first law of thermodynamics: $$ \Delta U = Q - W $$ where:

• $\Delta U = 0.25 , \text{kJ}$
• $W = 4 , \text{kJ}$

Rearranging to find heat transfer $Q$: $$ Q = \Delta U + W = 0.25 , \text{kJ} + 4 , \text{kJ} = 4.25 , \text{kJ} $$

3. Calculate Work for the Piston as the System

For the piston, the work done on the piston can be considered as the work done against the atmospheric pressure: $$ W_{piston} = P_{atm} \cdot \Delta V = 1 \text{ bar} \cdot (V_2 - V_1) $$ Substituting values: $$ W_{piston} = 100 , \text{kPa} \times 0.02 , \text{m}^3 = 2 , \text{kJ} $$

4. Evaluate Change in Potential Energy for the Piston

Since specific details about the height change are not given here, we assume no significant change in height which leads to: $$ \Delta PE = 0 , \text{kJ} $$

In a typical case, if height change were provided, we would use: $$ \Delta PE = m \cdot g \cdot h $$ But in this scenario, we state that potential energy change can be approximated as zero.