Multiple Choice
At what temperature in °C does 250.0 mL of argon gas at 25°C and 1.50 atm occupy a volume of 450.0 mL at a pressure of 1.90 atm, according to the Ideal Gas Law?
7. Gases
The Ideal Gas Law Applications
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What is the pressure in mm of Hg of a gas mixture that contains 1g of H2 and 8.0g of Ar in a 3.0 L container at 27°C?
A
720 mm of Hg
B
745 mm of Hg
C
780 mm of Hg
D
760 mm of Hg
Verified step by step guidance1
Convert the mass of each gas to moles using their molar masses. For H2, use the molar mass of approximately 2.02 g/mol, and for Ar, use 39.95 g/mol.
Calculate the number of moles of H2: \( n_{H2} = \frac{1 \text{ g}}{2.02 \text{ g/mol}} \). Calculate the number of moles of Ar: \( n_{Ar} = \frac{8.0 \text{ g}}{39.95 \text{ g/mol}} \).
Use the ideal gas law to find the total pressure of the gas mixture. The ideal gas law is \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is the total number of moles, \( R \) is the ideal gas constant (0.0821 L·atm/mol·K), and \( T \) is the temperature in Kelvin.
Convert the temperature from Celsius to Kelvin: \( T = 27 + 273.15 = 300.15 \text{ K} \).
Calculate the total pressure in atm using the ideal gas law: \( P = \frac{(n_{H2} + n_{Ar})RT}{V} \). Convert the pressure from atm to mm of Hg using the conversion factor 1 atm = 760 mm of Hg.
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