A 9.25-L container holds a mixture of two gases at 13 °C. Help PLEASE!?
2016-10-13 6:11 pm
A 9.25-L container holds a mixture of two gases at 13 °C. The partial pressures of gas A and gas B, respectively, are 0.401 atm and 0.577 atm. If 0.180 mol of a third gas is added with no change in volume or temperature, what will the total pressure become
回答 (1)
2016-10-13 6:30 pm
For the third gas :
Partial pressure = ? atm
Volume, V = 9.25 L
Number of moles, n = 0.180 mol
Gas constant, R = 0.0821 atm L / (mol K)
Absolute temperature, T = (273 + 13) K = 286 K
PV = nRT
Hence, P = nRT/V
Partial pressure of the third gas = 0.180 × 0.0821 × 286 / 9.25 atm = 0.457 atm
Total pressure = (0.401 + 0.577 + 0.457) atm = 1.435 atm
Partial pressure = ? atm
Volume, V = 9.25 L
Number of moles, n = 0.180 mol
Gas constant, R = 0.0821 atm L / (mol K)
Absolute temperature, T = (273 + 13) K = 286 K
PV = nRT
Hence, P = nRT/V
Partial pressure of the third gas = 0.180 × 0.0821 × 286 / 9.25 atm = 0.457 atm
Total pressure = (0.401 + 0.577 + 0.457) atm = 1.435 atm
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