Question

Container A holds 787 mL of ideal gas at 2.20 atm Container B holds 189 mL of ideal gas at 4.10 atm. If the gases are allowed to mix together, what is the partial pressure of each gas in the total volume? A gas A atm gas B: atm

Answer 1

We can find no. of moles of each gases prior to mixing.

We have equation P V = n R T i e n = P V / R T

where, P is a pressure of a gas, V is a volume of a gas, n is no. of moles of gas, R is a gas constant and T is temperature of gas.

$\therefore$ n _A = 2.20 atm $\times$ 0.787 L / RT

Similarly, n _B = 4.10 atm $\times$ 0.189 L / RT

After mixing two gases, total volume of gas is ( 0.787 L + 0.189 L ) 0.976 L.

Now, we can calculate partial pressures of both gases by using equation P = n R T / V.

P _A = ( 2.20 atm $\times$ 0.787 L / RT ) R T / 0.976 L = 1.77 atm

P _B= ( 4.10 atm $\times$ 0.189 L / RT ) R T / 0.976 L = 0.794 atm

ANSWER :

Partial pressure of gas A = 1.77 atm

Partial pressure of gas B = 0.794 atm