Gas is confined in a tank at a pressure of 15.5 atm and a temperature of 14.6◦C. If half of the gas is withdra?
Gas is confined in a tank at a pressure of
15.5 atm and a temperature of 14.6◦C.
If half of the gas is withdrawn and the
temperature is raised to 57.6◦C, what is the
new pressure in the tank?
Answer in units of atm.
HELP? I used the PV/T = PV/T formula but the right answer doesn't come out
Since half the gas came out
shouldnt the equation look like this?
15.5*1*/287.75 = x/(330.75*2)
and solve for x? I did this and the answer doesnt come out right
help pleaes
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Verified answer
PV = nRT
Solving for P and labeling the initial pressure as P'
P' = nRT/V
Final pressure is P"
Initial temperature is 287.6 K and final temperature is 330.6
P" = (n/2)R[(330.6/287.6)T]/V = (nRT/V) • (330.6/575.2)
P' = nRT/P = 15.5 atm
P" = (nRT/P) • (330.6/575.2) = 15.5 atm • (330.6/575.2) = 8.9 atm
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You shouldn't use the combined gas law because it doesn't include moles, you should instead use PV=nRT
That's why your answer wasn't coming out right, solving for x i got 35.6 atm, which is wrong.