# A sample of helium at 145 mmHg is in a 3-00 L container and a sample o.docx

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A sample of helium at 145 mmHg is in a 3.00 L container and a sample of argon at 355 mmHg is in a 2.00 L container, both at 28Â°C. If the two containers are connected and opened to each other at constant temperature, such that the total volume is now 5.00 L, what is the total pressure of the mixture of helium and argon? 355 mmHg 100 mmHg 145 mmHg 500 mmHg 229 mmHg Save Solution no of mol of He in sample(n) = PV/RT = ((145/760)*3/(0.0821*301)) = 0.0232 mol no of mol of Ar in sample(n) = PV/RT = ((355/760)*2/(0.0821*301)) = 0.0378 mol nTotal = 0.0232 + 0.0378 = 0.061 mol total volume = 3+2 = 5 L total pressure(pTotal) = nRT/V = 0.061*0.0821*301/5 = 0.301 atm = 0.301*760Â Â Â  ( 1 atm = 760 mmhg) = 229 mmhg answer: 229 mmhg .

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### A sample of helium at 145 mmHg is in a 3-00 L container and a sample o.docx

• 1. A sample of helium at 145 mmHg is in a 3.00 L container and a sample of argon at 355 mmHg is in a 2.00 L container, both at 28Â°C. If the two containers are connected and opened to each other at constant temperature, such that the total volume is now 5.00 L, what is the total pressure of the mixture of helium and argon? 355 mmHg 100 mmHg 145 mmHg 500 mmHg 229 mmHg Save Solution no of mol of He in sample(n) = PV/RT = ((145/760)*3/(0.0821*301)) = 0.0232 mol no of mol of Ar in sample(n) = PV/RT = ((355/760)*2/(0.0821*301)) = 0.0378 mol nTotal = 0.0232 + 0.0378 = 0.061 mol total volume = 3+2 = 5 L total pressure(pTotal) = nRT/V = 0.061*0.0821*301/5 = 0.301 atm = 0.301*760Â Â Â ( 1 atm = 760 mmhg) = 229 mmhg answer: 229 mmhg
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