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Chapter 10: Problem 44

Many gases are shipped in high-pressure containers. Consider a steel tank whose volume is 55.0 gallons that contains \(\mathrm{O}_{2}\) gas at a pressure of \(16,500 \mathrm{kPa}\) at \(23{ }^{\circ} \mathrm{C}\). (a) What mass of \(\mathrm{O}_{2}\) does the tank contain? (b) What volume would the gas occupy at STP? (c) At what temperature would the pressure in the tank equal 150.0 atm? (d) What would be the pressure of the gas, in \(\mathrm{kPa},\) if it were transferred to a container at \(24^{\circ} \mathrm{C}\) whose volume is \(55.0 \mathrm{~L} ?\)

Short Answer

Expert verified
The mass of \(\mathrm{O}_{2}\) in the tank is approximately 4662.17 g. At STP, the gas would occupy a volume of approximately 3267.93 L. The temperature at which the pressure in the tank equals 150.0 atm is approximately 324.60 K. The pressure of the gas in the new container is approximately 75207.33 kPa.

Step by step solution

01

Apply the Ideal Gas Law to find moles of \(\mathrm{O}_{2}\)

Using the Ideal Gas Law formula, \(PV=nRT\), solve for the number of moles (n) in the tank: \(n = \frac{PV}{RT}\) We can then plug in our given values: n = \(\frac{162.861 \text{ atm} \cdot 208.197 \text{ L}}{(0.0821(\text{L}\cdot\text{atm})/(\text{mol}\cdot\text{K})) \cdot 296.15 \text{ K}}\) n ≈ 145.818 mol
02

Calculate the mass of \(\mathrm{O}_{2}\)

We know the number of moles (n) and the molar mass of \(\mathrm{O}_{2}\). Now, the mass (m) can be calculated using the formula: mass of \(\mathrm{O}_{2}\) (m) = n × molar mass of \(\mathrm{O}_{2}\) m = 145.818 mol × 32.00 g/mol m ≈ 4662.17 g The mass of \(\mathrm{O}_{2}\) in the tank is approximately 4662.17 g. b) Volume the gas would occupy at STP (STP: standard temperature and pressure; P=1 atm, T=273.15 K)
03

Calculate the volume at STP

Using the Ideal Gas Law formula, \(PV=nRT\), solve for the volume (V) at STP: V = \(\frac{nRT}{P}\) We can then plug in the values at STP: V = \(\frac{145.818 \text{ mol} \cdot (0.0821(\text{L}\cdot\text{atm})/(\text{mol}\cdot\text{K})) \cdot 273.15 \text{ K}}{1 \text{ atm}}\) V ≈ 3267.93 L At STP, the gas would occupy a volume of approximately 3267.93 L. c) Temperature at which the pressure in the tank equals 150.0 atm
04

Calculate the temperature

Using the Ideal Gas Law formula, \(PV=nRT\), solve for the temperature (T) when pressure is 150.0 atm: T = \(\frac{PV}{nR}\) We can then plug in the values: T = \(\frac{150.0 \text{ atm} \cdot 208.197 \text{ L}}{145.818 \text{ mol} \cdot (0.0821(\text{L}\cdot\text{atm})/(\text{mol}\cdot\text{K}))}\) T ≈ 324.60 K The temperature at which the pressure in the tank equals 150.0 atm is approximately 324.60 K. d) Pressure of the gas in a 55.0 L container at 24°C Temperature at 24°C = 297.15 K
05

Calculate the new pressure

Using the Ideal Gas Law formula, \(PV=nRT\), solve for the pressure (P) in the new container: P = \(\frac{nRT}{V}\) We can then plug in the values: P = \(\frac{145.818 \text{ mol} \cdot (0.0821(\text{L}\cdot\text{atm})/(\text{mol}\cdot\text{K})) \cdot 297.15 \text{ K}}{55.0 \text{ L}}\) P ≈ 742.431 atm Now, let's convert this pressure to kPa: P = 742.431 atm × 101.325 kPa/atm P ≈ 75207.33 kPa The pressure of the gas in the new container is approximately 75207.33 kPa.

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