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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?

Respuesta :

Answer:

The total  pressure P = 1.642 atm

Explanation:

From the dalton's law

Total pressure of the mixture is

[tex]P = P_{A} + P_{B} + P_{C}[/tex] ----- (1)

[tex]P_A = 0.419[/tex] atm

[tex]P_B = 0.589[/tex] atm

[tex]P_C = \frac{nRT}{V}[/tex]

n = 0.24 mole

R = [tex]0.0821[/tex] [tex]\frac{L.Atm}{K.mol}[/tex]

T = 41 °c = 314 K

[tex]P_C = \frac{(0.24)(0.0821)(314)}{9.75}[/tex]

[tex]P_C[/tex] = 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

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