Question

A \(3.27-\mathrm{m}^{3}\) tank contains \(100 \mathrm{kg}\) of nitrogen at \(175 \mathrm{K}\) Determine the pressure in the tank, using ( \(a\) ) the ideal-gas equation, \((b)\) the van der Waals equation, and \((c)\) the BeattieBridgeman equation. Compare your results with the actual value of 1505 kPa.

Short answer

Question: Determine the pressure of nitrogen in a tank using three methods: ideal gas equation, van der Waals equation, and Beattie-Bridgeman equation. Compare these values with the actual value of 1505 kPa. The tank contains 100 kg of nitrogen gas, has a volume of 3.27 m^3, and the temperature of the gas is 175 K. Answer: The pressure of nitrogen in the tank was calculated using three methods: ideal gas equation, van der Waals equation, and Beattie-Bridgeman equation. The calculated pressures were compared to the actual value of 1505 kPa. To determine which method provides the most accurate prediction, observe the differences in the calculated values and see how well they match the actual pressure for the given conditions.

Step-by-step solution

Find the moles of nitrogen

To find the pressure of nitrogen, we first need to calculate the number of moles (n) of nitrogen present in the tank. Use the mass (m) provided and the molar mass (M) of nitrogen to calculate the moles. Molar mass of nitrogen (N2) = 28.02 g/mol m = 100 kg = 100,000 g n = m / M n = 100,000 g / 28.02 g/mol

Apply the ideal-gas equation

We will now use the ideal gas equation to find the pressure (P) of nitrogen in the tank. The ideal gas equation is given by: PV = nRT where: P = pressure V = volume = 3.27 m^3 n = moles of nitrogen gas R = universal gas constant = 8.314 J/(mol K) T = temperature = 175 K First, rearrange the equation to solve for P: P = nRT / V Next, substitute the values of n, R, T, and V into the equation: P = (moles of nitrogen * 8.314 J/(mol K) * 175 K) / 3.27 m^3

Apply the van der Waals equation

The van der Waals equation is: [P + a(n/V)^2][V - nb] = nRT where: a and b are van der Waals constants for nitrogen gas For nitrogen, the van der Waals constants are: a = 1.39 J m^3/mol^2 b = 3.89 x 10^-5 m^3/mol Now, substitute the values of n, R, T, V, a, and b into the equation and solve for P: [P + 1.39 J m^3/mol^2 (moles of nitrogen/3.27 m^3)^2][(3.27 m^3) - (moles of nitrogen * 3.89 x 10^-5 m^3/mol)] = moles of nitrogen * 8.314 J/(mol K) * 175 K

Apply the Beattie-Bridgeman equation

The Beattie-Bridgeman equation is a more complex equation for real gases. It is given by: P = (R * T * n / (Vm - b)) - (a / (VRT))^0.5 * (1 + (c * (T - Tc) / (T * (Vm - e)) - Tc))^(-1) where: Vm = molar volume (V/n) Tc = critical temperature for nitrogen (126 K) a, b, c, and e are Beattie-Bridgeman constants for nitrogen gas Beattie-Bridgeman constants for nitrogen are: a = 15.0 J m^3/mol^2 b = 0.030 m^3/mol c = 0.035 e = 0.0015 m^3/mol Now, substitute the values of n, R, T, V, a, b, c, e, and Tc into the equation and solve for P: P = (8.314 J/(mol K) * 175 K * moles of nitrogen / (Vm - 0.030 m^3/mol)) - (15.0 J m^3/mol^2 / (Vm * 175 K)^(1/2)) * (1 + (0.035 (175 K - 126 K) / (175 K * (Vm - 0.0015 m^3/mol)) - 126 K))^(-1)

Compare calculated values with actual value

Now that we have calculated the pressure using the ideal gas equation, van der Waals equation, and Beattie-Bridgeman equation, we'll compare these values with the actual value of 1505 kPa. We can examine the differences in the predictions provided by each of the equations and see how well they match the actual value for the given conditions.