Question

how many moles of methane occupy a volume of 2.00 l at 50.0 °c and 0.500 atm? (r = 0.0821 atm·l/mol·k)

0.0377 mol
26.5 mol
4.11 mol
0.244 mol
0.151 mol

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, and \( T \) is temperature in Kelvin. We need to solve for \( n \), so rearrange the formula to \( n=\frac{PV}{RT} \).

Step2: Convert Temperature to Kelvin

Given \( T = 50.0^\circ\text{C} \), convert to Kelvin: \( T = 50.0 + 273.15 = 323.15\,\text{K} \).

Step3: Substitute Values into Formula

We have \( P = 0.500\,\text{atm} \), \( V = 2.00\,\text{L} \), \( R = 0.0821\,\text{atm·L/mol·K} \), and \( T = 323.15\,\text{K} \). Substitute into \( n=\frac{PV}{RT} \):
\( n=\frac{(0.500\,\text{atm})(2.00\,\text{L})}{(0.0821\,\text{atm·L/mol·K})(323.15\,\text{K})} \).

Step4: Calculate Numerator and Denominator

Numerator: \( (0.500)(2.00) = 1.00\,\text{atm·L} \).
Denominator: \( (0.0821)(323.15) \approx 26.53\,\text{atm·L/mol} \).
Then \( n=\frac{1.00}{26.53} \approx 0.0377\,\text{mol} \).

Answer:

0.0377 mol