Asked by Anonymous
Calculate the partial pressures and total pressure (in atm) exerted by 0.838 mol of methane, 0.184 mol of ethane and 0.755 mol of neon contained in a 4.00 L container at 27 C.
PV=nRT for methane
(4.00L) X = (0.838 mol)(0.08206 L-atm/mol-K)(300.15 K)
X = 5.160 atm
PV=nRT for ethane
(4.00L) X = (0.184 mol)(0.08206 L-atm/mol-K)(300.15 K)
X = 1.133 atm
PV=nRT for neon
(4.00L) X = (0.755 mol)(0.08206 L-atm/mol-K)(300.15 K)
X = 4.649 atm
Total Pressure: 5.160atm + 1.133atm + 4.649atm = 10.942 atm
Correct?
PV=nRT for methane
(4.00L) X = (0.838 mol)(0.08206 L-atm/mol-K)(300.15 K)
X = 5.160 atm
PV=nRT for ethane
(4.00L) X = (0.184 mol)(0.08206 L-atm/mol-K)(300.15 K)
X = 1.133 atm
PV=nRT for neon
(4.00L) X = (0.755 mol)(0.08206 L-atm/mol-K)(300.15 K)
X = 4.649 atm
Total Pressure: 5.160atm + 1.133atm + 4.649atm = 10.942 atm
Correct?
Answers
Answered by
DrBob222
The numbers are ok and the method is correct; however, you have too many significant figures in the answer (for each and for the total).
Answered by
Anonymous
So, for methane, would it be 5.16 atm; ethane 1.13 atm; neon 4.65 atm; giving a total pressure of 10.94 atm?
Answered by
DrBob222
That's right.
Answered by
Anonymous
Great, thanks
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