Question

A gaseous mixture made from 12.08 g O₂, 10.22 g CH₄, and 12.32 g of CO₂ is placed in a 18.34 L vessel at 2 °C. What is the total pressure in the vessel?

Answer 1

Moles of O2 = n1 = mass/molar mass = 12.08/32 = 0.3775

Similarly,

Moles of CH4 = n2 = 10.22/16 = 0.63875

Moles of CO2 = n3 = 12.32/44 = 0.28

Temperature (T) = 2°C = 2 + 273 = 275 K

Volume of vessel (V) = 18.34 liters

Gas constant (R) = 0.0821 L atm/mol K

Now, we will calculate pressure of individual gases one by one by using ideal gas equation.

1) Let pressure of O2 is P1 atm

Then,

P1 * V = n1 * R * T

P1 = n1 * R * T/V = 0.3775 * 0.0821 * 275 / 18.34 = 0.4647 atm

2) Let P2 is pressure of CH4

P2 = n2 * R * T/V = 0.63875*0.0821*275 / 18.34 = 0.8417 atm

3) P3 = n3 * R * T / V = 0.28 * 0.0821 * 275 / 18.34 = 0.3447 atm

Total pressure (P) = P1 + P2 + P3 = 0.4647+0.8417+0.3447 = 1.65 atm .... Answer

Let me know if any doubts.