Suppose that 120.mL of argon is collected over water at 25C and 780.0 torr. Compute the volume of the dry argon at STP.
Would this be involved with Charles law?
DrBob11:
(P1V1/T1) = (P2V2/T2)
Don't forget to use T in Kelvin. Subtract the vapor pressure of water from the total pressure for this problem as well as the previous one.
Why do I have to subtract the vapor pressure of water? Shouldn't I be ignoring water vapor because the answer is talking about dry Ar though??
We have never learned the formula
(P1V1/T1) = (P2V2/T2)
Is that the same thing as
P1V1T2=P2V2T1 ??
2 answers
Sure.
(P1V1/T1) = (P2V2/T2) =
P1V1T2 = P2V2T2.
That's the general gas law. It becomes Boyle's Law if you keep T constant (just cover up T1 and T2 and you're left with P1V1=P2V2).
Or it becomes Charle's Law if P is constant (cover up P1 and P2 and it becomes V1T2=V2T1).
And if V is held constant, (cover V1 and V2) and you have P1T2 = P2T1.
(P1V1/T1) = (P2V2/T2) =
P1V1T2 = P2V2T2.
That's the general gas law. It becomes Boyle's Law if you keep T constant (just cover up T1 and T2 and you're left with P1V1=P2V2).
Or it becomes Charle's Law if P is constant (cover up P1 and P2 and it becomes V1T2=V2T1).
And if V is held constant, (cover V1 and V2) and you have P1T2 = P2T1.
1 answer