A 8.50 L container holds a mixture of two gases at 11°C. The partial pressure of

Question

A 8.50 L container holds a mixture of two gases at 11°C. The partial pressure of gas A and gas B, respectively, are 0.294 atm and 0.816atm. If 0.250 mil of a third gas is added with no change in volume or temperature, what will the total pressure become? A 8.50 L container holds a mixture of two gases at 11°C. The partial pressure of gas A and gas B, respectively, are 0.294 atm and 0.816atm. If 0.250 mil of a third gas is added with no change in volume or temperature, what will the total pressure become?

Explanation / Answer

Total pressure of gases mixture (P) = PA + PB + PC

nc   = 0.25moles

T   =11+273 = 284K

V = 8.5L

PC   = ncRT/V

        = 0.25*0.0821*284/8.5   = 0.686atm

PA   = 0.294atm

PB   = 0.816atm

Total pressure of the gases mixture = PA + PB + PC

                                                           = 0.294+0.816+0.686   = 1.796atm >>>>>answer

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