Question

A 9.75 L 9.75 L container holds a mixture of two gases at 41 ° C. 41 °C. The partial pressures of gas A and gas B, respectively, are 0.419 atm 0.419 atm and 0.589 atm. 0.589 atm. If 0.240 mol 0.240 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?

Answer 1

Answer:

The total  pressure P = 1.642 atm

Explanation:

From the dalton's law

Total pressure of the mixture is

$P = P_{A} + P_{B} + P_{C}$ ----- (1)

$P_A = 0.419$ atm

$P_B = 0.589$ atm

$P_C = \frac{nRT}{V}$

n = 0.24 mole

R = $0.0821$ $\frac{L.Atm}{K.mol}$

T = 41 °c = 314 K

$P_C = \frac{(0.24)(0.0821)(314)}{9.75}$

$P_C$ = 0.634 atm

From equation (1)

P = 0.419 + 0.589 + 0.634

P = 1.642 atm

Thus the total  pressure P = 1.642 atm