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Chapter 8: Problem 144

A steel cylinder contains \(150.0\) moles of argon gas at a temperature of \(25^{\circ} \mathrm{C}\) and a pressure of \(8.93 \mathrm{MPa}\). After some argon has been used, the pressure is \(2.00 \mathrm{MPa}\) at a temperature of \(19^{\circ} \mathrm{C}\). What mass of argon remains in the cylinder?

Short Answer

Expert verified
The mass of argon remaining in the cylinder is approximately \(2536 \text{ g}\).

Step by step solution

01

Identify the Ideal Gas Law formula

The Ideal Gas Law formula is given by: PV = nRT Where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature in Kelvin.
02

Convert the given temperatures to Kelvin

The temperatures given are in Celsius, so we need to convert them to Kelvin: Initial temperature: \(T_1 = 25^{\circ}C + 273.15 = 298.15 K\) Final temperature: \(T_2 = 19^{\circ}C + 273.15 = 292.15 K\)
03

Relate the initial and final conditions using the Ideal Gas Law

Since the volume of the cylinder is constant and we are given \(n_1, P_1, T_1, P_2,\) and \(T_2\), we can relate the initial and final conditions as follows: \(\frac{P_1 V}{T_1} = n_1 R \Rightarrow V = \frac{n_1 RT_1}{P_1}\) Similarly, for the final conditions: \(\frac{P_2 V}{T_2} = n_2 R \Rightarrow V = \frac{n_2 RT_2}{P_2}\) Since the volume is constant, the two expressions for V should be equal: \(\frac{n_1 RT_1}{P_1} = \frac{n_2 RT_2}{P_2}\)
04

Solve for the final number of moles of argon, \(n_2\)

We can now solve for the final number of moles of argon, \(n_2\). The gas constant R will cancel out from both sides: \(n_2 = \frac{n_1 P_2 T_1}{P_1 T_2}\) Substitute the given values and solve for \(n_2\): \(n_2 = \frac{150.0 \text{ moles} \times 2.00 \text{ MPa} \times 298.15 \text{ K}}{8.93 \text{ MPa} \times 292.15 \text{ K}} \approx 63.47 \text{ moles}\)
05

Calculate the mass of argon remaining in the cylinder

Now that we have the final number of moles of argon, we can calculate the mass of argon remaining in the cylinder using the molar mass of argon: Molar mass of argon: \(39.95 \frac{\text{g}}{\text{mole}}\) Mass of argon remaining: \(m_2 = n_2 \times \text{Molar mass of argon}\) \(m_2 = 63.47 \text{ moles} \times 39.95 \frac{\text{g}}{\text{mole}} \approx 2536 \text{g}\) Therefore, the mass of argon remaining in the cylinder is approximately 2536 grams.

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Most popular questions from this chapter

A \(15.0-\mathrm{L}\) tank is filled with \(\mathrm{H}_{2}\) to a pressure of \(2.00 \times 10^{2}\) atm. How many balloons (each \(2.00 \mathrm{L}\) ) can be inflated to a pressure of 1.00 atm from the tank? Assume that there is no temperature change and that the tank cannot be emptied below \(1.00\) atm pressure.

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An important process for the production of acrylonitrile \(\left(\mathrm{C}_{3} \mathrm{H}_{3} \mathrm{N}\right)\) is given by the following equation: $$2 \mathrm{C}_{3} \mathrm{H}_{6}(g)+2 \mathrm{NH}_{3}(g)+3 \mathrm{O}_{2}(g) \longrightarrow 2 \mathrm{C}_{3} \mathrm{H}_{3} \mathrm{N}(g)+6 \mathrm{H}_{2} \mathrm{O}(g)$$ A \(150 .\)-\(\mathrm{L}\)reactor is charged to the following partial pressures at \(25^{\circ} \mathrm{C}:\) $$\begin{aligned}P_{\mathrm{C}, \mathrm{H}_{6}} &=0.500 \mathrm{MPa} \\\P_{\mathrm{NH}_{3}} &=0.800 \mathrm{MPa} \\\P_{\mathrm{O}_{2}} &=1.500 \mathrm{MPa}\end{aligned}$$ What mass of acrylonitrile can be produced from this mixture \(\left(\mathrm{MPa}=10^{6} \mathrm{Pa}\right) ?\)

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