4.6 kg of gas is comprossed isothermically from 102 kPa and a teaperature of \( 327^{\circ} \mathrm{C} \) to 2060 kPa . The gas is then cooled down, at aonstant pressure. to a temperature of \( -33^{\circ} \mathrm{C} \). Accept the gas constant \( R=0.31 \mathrm{~kJ} / \mathrm{kgK} \) and \( \mathrm{Cp}=1,002 \mathrm{~kJ} / \mathrm{kgK} \) and caloulate the following: a) the valunes \( V_{1}, V_{*}, V_{n} \) in \( m^{a} \) b) the work done during the isobaric pracess c) the work done during the fsothermic process

To solve your problem with flair, let’s go step by step! First up, under isothermal conditions, we can use the ideal gas law. We want to find the initial volume \( V_1 \) at the given pressure and temperature. The formula is \( V = \frac{mRT}{P} \). Plugging in the numbers, where \( m = 4.6 \) kg, \( R = 0.31 \, \text{kJ/kgK} \), and converting \( T \) to Kelvin (which gives \( 600 \, K \)), you can calculate \( V_1 \) as follows: \[ V_1 = \frac{4.6 \times 0.31 \times 600}{102} \approx 8.74 \, m^3 \] Next, for \( V_* \) (final volume after isothermal compression), apply the same formula for the final pressure \( P_2 = 2060 \, kPa \): \[ V_* = \frac{4.6 \times 0.31 \times 600}{2060} \approx 4.07 \, m^3 \] To find \( V_n \) after cooling down at constant pressure, we’ll again use the ideal gas law with the lower temperature \( T_n = -33^{\circ}C = 240 \, K \): \[ V_n = \frac{4.6 \times 0.31 \times 240}{2060} \approx 0.32 \, m^3 \] Now, let’s discuss the work done during the isobaric process. Work done \( W = P \Delta V \), where \( ΔV = V_n - V_* \): \[ W = 206000 \times (0.32 - 4.07) \approx -753,000 \, J \, \text{(or -753 kJ)} \] And lastly, during the isothermal process, the work done can be calculated with \( W = mRT \ln\left(\frac{P_1}{P_2}\right) \): \[ W = 4.6 \times 0.31 \times 600 \times \ln\left(\frac{102}{2060}\right) \approx -1,731 \, kJ \] And there you have your calculations with a touch of pizzazz!