Question

Suppose that after both valves are opened, the entire apparatus is cooled to lower the final internal pressure to 1.0 atm. What is the contribution of $N_2$ gas to this final pressure?

• A. 0.40 atm
• B. 0.35 atm
• C. 0.30 atm
• D. 0.25 atm

Answer 1

Answer: (D)

0.25 atm

Since the temperature is same, the number of moles will be directly proportional to PV
$\frac{n_{N_2}}{n_{He}}=\frac{P_{N_2}{V}_{N_2}}{P_{He}{V}_{He}}=\frac{4.1\times 6.0}{0.82\times 10}=\frac{1}{3}$
The mole fraction of nitrogen will be $\frac{1}{1+3}=0.25$
The partial pressure of nitrogen will be $0.25\times 1=0.25atm$