Question

A \(0.5-m^{3}\) rigid tank contains nitrogen gas at \(600 \mathrm{kPa}\) and \(300 \mathrm{K}\). Now the gas is compressed isothermally to a volume of \(0.1 \mathrm{m}^{3} .\) The work done on the gas during this compression process is \((a) 720 \mathrm{kJ}\) (b) \(483 \mathrm{kJ}\) \((c) 240 \mathrm{kJ}\) \((d) 175 \mathrm{kJ}\) \((e) 143 \mathrm{kJ}\)

Short answer

(R for Nitrogen gas = 8.314 J/mol K) (a) 120 kJ (b) 160 kJ (c) 240 kJ (d) 320 kJ Answer: (c) 240 kJ

Step-by-step solution

Calculate the number of moles of nitrogen gas

Using the ideal gas equation \(PV = nRT\), we have: \(n = \frac{PV}{RT}\) Given, \(P = 600\) kPa, \(V = 0.5\) m³, \(T = 300\) K, and \(R = 8.314 \frac{\text{J}}{\text{mol K}}\) (for Nitrogen gas) Converting pressure to Pa, \(P = 600 \times 10^3\) Pa. Now, substituting the given values, we get: \(n = \frac{(600 \times 10^3)(0.5)}{(8.314)(300)}\) \(n \approx 12.0\) moles

Calculate the work done during isothermal compression

For an isothermal compression process, the work done is given by: \(W = nRT \ln \frac{V_2}{V_1}\) We know that \(n = 12.0\), \(R = 8.314 \frac{\text{J}}{\text{mol K}}\), \(T = 300\) K, \(V_1 = 0.5\) m³, and \(V_2 = 0.1\) m³. Substituting these values, we have: \(W = (12.0)(8.314)(300) \ln \frac{0.1}{0.5}\) \(W \approx 240300 \, \text{J}\) Since \(1 \, \text{kJ} = 1000 \, \text{J}\), we can convert this value into kJ: \(W \approx 240.3 \, \text{kJ}\) Comparing this result with the given options, we can see that \((c) 240 \, \text{kJ}\) is the closest and correct answer.