What pressure would a gas mixture in a 10.0 L tank exert if it were composed of 48.5 g of He and 94.6 g of CO2 at 398 K, assuming ideal gas behavior?

• A. 12.3 atm
• B. 18.4 atm
• C. 15.8 atm
• D. 9.7 atm

Answer: (C)

Verified step by step guidance

1. First, calculate the number of moles of each gas using the formula: \( n = \frac{m}{M} \), where \( m \) is the mass of the gas and \( M \) is the molar mass. For helium (He), the molar mass is approximately 4.00 g/mol, and for carbon dioxide (CO2), it is approximately 44.01 g/mol.
2. Calculate the moles of helium: \( n_{\text{He}} = \frac{48.5 \text{ g}}{4.00 \text{ g/mol}} \).
3. Calculate the moles of carbon dioxide: \( n_{\text{CO2}} = \frac{94.6 \text{ g}}{44.01 \text{ g/mol}} \).
4. Use the ideal gas law to find the total pressure exerted by the gas mixture. The ideal gas law is given by \( PV = nRT \), where \( P \) is the pressure, \( V \) is the 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.
5. Calculate the total moles of gas by adding the moles of helium and carbon dioxide. Then, rearrange the ideal gas law to solve for pressure: \( P = \frac{nRT}{V} \). Substitute the total moles, the given temperature (398 K), the volume (10.0 L), and the ideal gas constant into the equation to find the pressure.