Question

A \(3-m^{3}\) rigid vessel contains steam at 4 MPa and \(500^{\circ} \mathrm{C} .\) The mass of the steam is \((a) 3 \mathrm{kg}\) \((b) 9 \mathrm{kg}\) \((c) 26 \mathrm{kg}\) \((d) 35 \mathrm{kg}\) \((e) 52 \mathrm{kg}\)

Short answer

a) 25 kg b) 30 kg c) 32 kg d) 35 kg Answer: d) 35 kg

Step-by-step solution

Convert units

Since the given units are in MPa and Celsius, we need to convert the pressure to kPa and the temperature to Kelvin, so that we can use the specific gas constant for steam in consistent units. - Pressure: 4 MPa = 4 * 1000 kPa = 4000 kPa - Temperature: \(500^{\circ} \mathrm{C}\) = 500 + 273.15 = 773.15 K

Rearrange ideal gas law formula for mass

Using P * V = m * R * T, we can rearrange the formula to calculate mass (m): m = (P * V) / (R * T)

Substitute the values and solve for mass

Now we can substitute the given values of P, V, R, and T into the formula and solve for the mass: m = (4000 kPa * 3 \(m^3\)) / (0.4615 kJ/kg K * 773.15 K) m ≈ (12000 kPa * \(m^3\)) / (355.689 kJ/kg) m ≈ 33.75 kg

Choose the closest option

Since the calculated mass is approximately 33.75 kg, the closest answer is (d) 35 kg. So, the mass of the steam is most likely around 35 kg.