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 28°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*760    ( 1 atm = 760 mmhg) = 229 mmhg answer: 229 mmhg .

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 28°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
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 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 28°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*760   ( 1 atm = 760 mmhg) = 229 mmhg answer: 229 mmhg