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Chapter 5: Problem 53

Suppose two \(200.0-\mathrm{L}\) tanks are to be filled separately with the gases helium and hydrogen. What mass of each gas is needed to produce a pressure of 2.70 atm in its respective tank at \(24^{\circ} \mathrm{C} ?\)

Short Answer

Expert verified
To produce a pressure of 2.70 atm in their respective tanks at \(24^{\circ} \mathrm{C}\), approximately \(88.00 \: \text{g}\) of helium and \(44.44 \: \text{g}\) of hydrogen are needed.

Step by step solution

01

Convert the temperature from Celsius to Kelvin

Before we begin, we need to convert the temperature from Celsius to Kelvin since the Ideal Gas Law requires the temperature to be in Kelvin. The conversion formula is: Temperature (K) = Temperature (°C) + 273.15 For our problem, our temperature in Kelvin will be: Temperature (K) = 24°C + 273.15 = 297.15 K
02

Find the number of moles of each gas using the Ideal Gas Law

Using the Ideal Gas Law formula (PV = nRT), we can find the number of moles (n) of each gas by rearranging the formula: n = PV / RT For both gases, we are given: - Pressure (P) = 2.70 atm - Volume (V) = 200.0 L - Temperature (T) = 297.15 K - Ideal Gas Constant (R) = 0.0821 L atm/mol K Now, we can plug these values into the Ideal Gas Law formula: n(He) = n(H₂) = (2.70 atm) × (200.0 L) / (0.0821 L atm/mol K × 297.15 K) n(He) = n(H₂) ≈ 22.00 mol
03

Convert the moles of each gas to mass

The mass of a gas can be found by multiplying the number of moles of the gas by its molar mass. The molar mass of helium (He) is 4.00 g/mol, while the molar mass of hydrogen (H₂) is 2.02 g/mol. Mass (He) = Moles (He) × Molar Mass (He) Mass (He) ≈ 22.00 mol × 4.00 g/mol = 88.00 g Mass (H₂) = Moles (H₂) × Molar Mass (H₂) Mass (H₂) ≈ 22.00 mol × 2.02 g/mol ≈ 44.44 g Thus, the mass needed for each gas is approximately 88.00 g of helium and 44.44 g of hydrogen to produce a pressure of 2.70 atm in their respective tanks at 24°C.

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