Question

1. what volume is occupied by 0.118 mol of helium gas at a pres- sure of 0.97 atm and a temperature of 305 k? would the volume be different if the gas was argon (under the same conditions)?

Explanation:

Step1: Recall Ideal Gas Law

The Ideal Gas Law is \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is moles, \( R \) is gas constant, \( T \) is temperature. We solve for \( V \): \( V=\frac{nRT}{P} \). \( R = 0.0821\space L\cdot atm/(mol\cdot K) \).

Step2: Plug in values for Helium

Given \( n = 0.118\space mol \), \( P = 0.97\space atm \), \( T = 305\space K \), \( R = 0.0821\space L\cdot atm/(mol\cdot K) \).
\( V=\frac{0.118\space mol\times0.0821\space L\cdot atm/(mol\cdot K)\times305\space K}{0.97\space atm} \).
Calculate numerator: \( 0.118\times0.0821\times305 \approx 0.118\times25.0405 \approx 2.9548 \).
Divide by \( 0.97 \): \( V \approx \frac{2.9548}{0.97} \approx 3.056\space L \approx 3.06\space L \).

Step3: Volume for Argon (Ideal Gas Law)

Argon is a noble gas, behaves ideally under these conditions. Molar volume depends on \( n, P, T \), not identity (ideal gas). So \( V \) for argon (same \( n, P, T \)) is same as helium: \( \approx 3.06\space L \).

Answer:

Volume of helium: \(\approx 3.06\space L\); Volume of argon is same (since ideal gas, \( V \) depends on \( n, P, T \), not gas identity).