Answer: b. The partial pressure of N₂ is 202 kPa.
Explanation:
According to the ideal gas equation:'
\(PV=nRT\)
Pressure of the gas = ?
Volume of the gas = 11.2 L
Temperature of the gas = 0°C = 273K(0°C = 273 K)
Value of gas constant R=8.314 kPaL\Kmol
n = moles of gas = 3.0
\(P=\frac{nRT}{V}=\frac{3\times 8.314\times 273}{11.2}=607kPa\)
\(p_{N_2}=x_{N_2}\times P\)
\(x_{N_2}\) = mole fraction= \(\frac{\text {moles of}{N_2}}{\text {total moles}}=\frac{1}{3}=0.33\)
\(p_{N_2}=0.33\times 607=202kPa\)
\(p_{H_2}=x_{H_2}\times P\)
\(x_{H_2}\) = mole fraction= \(\frac{\text {moles of}{H_2}}{\text {total moles}}=\frac{2}{3}=0.67\)
\(p_{H_2}=0.67\times 607=405kPa\)
