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CHEM HELP!!!!?
A 9.30-L container holds a mixture of two gases at 25 °C. The partial pressures of gas A and gas B, respectively, are 0.175 atm and 0.688 atm. If 0.190 mol of a third gas is added with no change in volume or temperature, what will the total pressure become?
1 Answer
- Roger the MoleLv 73 years ago
0.175 atm + 0.688 atm = 0.863 atm total A & B
n = PV / RT = (0.863 atm) x (9.30 L) / ((0.08205746 L atm/K mol) x (25 + 273) K) = 0.3282 mol total A & B
(0.863 atm) x (0.3282 mol + 0.190 mol) / (0.3282 mol) = 1.36 atm total three gases
