A 9.75-L container holds a mixture of two gases at 27 °C. The partial pressures of gas A and gas B, respectively, are 0.428 atm and 0.705 atm. If 0.230 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?
I got 7.641 which was marked wrong
Please help Asap!?
2016-12-13 9:08 am
回答 (1)
2016-12-13 9:42 am
✔ 最佳答案
Consider the third gas :Partial pressure, P = ? atm
Volume, V = 9.75 L
No. of moles, n = 0.230 mol
Gas constant, R = 0.0821 L atm / (mol K)
Absolute temperature, T = (273 + 27) K = 300 K
PV = nRT
Then, P = nRT/V
Partial pressure of the third gas, P = 0.230 × 0.0821 × 300 / 9.75 atm = 0.581 atm
Total pressure = Sum of partial pressure = (0.428 + 0.705 + 0.581) atm = 1.714 atm
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