3. The isothermal compression of 0.3m^3 of air with initial parameters p_(1)=10 bar and t_(1)= 300^circ C produces 500 kJ of heat Determine the final volume V_(2) and the final pressure p_(2)

To solve this problem, we can use the first law of thermodynamics and the ideal gas law.<br /><br />Given information:<br />- Initial volume: $V_1 = 0.3 \, m^3$<br />- Initial pressure: $p_1 = 10 \, bar$<br />- Initial temperature: $T_1 = 300^{\circ}C$<br />- Heat added: $Q = 500 \, kJ$<br /><br />Step 1: Convert the initial temperature to Kelvin.<br />$T_1 = 300^{\circ}C + 273.15 = 573.15 \, K$<br /><br />Step 2: Use the first law of thermodynamics to find the change in internal energy.<br />$\Delta U = Q - W$<br />where $W$ is the work done by the system.<br /><br />Step 3: Use the ideal gas law to find the final volume and pressure.<br />$PV = nRT$<br />where $P$ is the pressure, $V$ is the volume, $n$ is the number of moles, $R$ is the specific gas constant, and $T$ is the temperature.<br /><br />Since the process is isothermal, the temperature remains constant, and the change in internal energy is zero.<br />$\Delta U = 0$<br /><br />Substituting the values, we get:<br />$0 = 500 \, kJ - W$<br /><br />Solving for $W$, we get:<br />$W = 500 \, kJ$<br /><br />Now, we can use the ideal gas law to find the final volume and pressure.<br /><br />$P_1V_1 = nRT_1$<br />$P_2V_2 = nRT_2$<br /><br />Since the number of moles and the gas constant remain constant, we can divide the two equations to eliminate $n$ and $R$:<br />$\frac{P_1V_1}{T_1} = \frac{P_2V_2}{T_2}$<br /><br />Substituting the known values, we get:<br />$\frac{10 \, bar \times 0.3 \, m^3}{573.15 \, K} = \frac{P_2 \times 0.3 \, m^3}{573.15 \, K}$<br /><br />Solving for $P_2$, we get:<br />$P_2 = 10 \, bar$<br /><br />Therefore, the final volume $V_2$ is the same as the initial volume, which is $0.3 \, m^3$, and the final pressure $P_2$ is $10 \, bar$.