Chemistry Gas Laws?

Question

An isolated quantity of an ideal gas at 307.4 K has a volume of 16.93 L at a pressure of 1.49 atm.
What is the absolute temperature of this gas sample when the pressure is halved and the volume is divided by four?
Specify temperature units as "K".

Answer 1

PV=nRT

P=Pressure
V=Volume
n=# moles
R= .0821 L*atm/mol*K
T= Temperature in K

Since it's an isolated quantity, amount of gas (n) won't change, none will get in or out.


So lets say you have no numbers at all. You just know n stays the same, P2=0.5P1, and V2=0.25V1

Use the combined gas law:
P1V1/T1=P2V2/T2
PV/T1=0.5P*0.25V/T2
PV/T1=0.125PV/T2
1/T1=0.125/T2
T2= 0.125*T1
T2=0.125(307.4)=38.4K

Or the long way:
To start:
P=1.49atm
V=16.93L
T=307.4K

Solve for n:
n=PV/RT
n=9.88x10^-3 mol

For the second part:
P=0.745atm
V=4.2325L
n=n from part 1=9.88x10^-3mol

Solve for T:
T=PV/nR=38.4K, which is 0.125T1

Answer 2

You somewhat use PV=nRT once you're given 3 of the 4 variables, tension, quantity, temp and/or moles of X (n). while you're in ordinary terms given tension and quantity, case in point, you need to use P1V1=P2V2 and resolve for despite you have no longer have been given. comparable with V1T1=V2V2, or perhaps P1V1T1 = P2V2T2 do you already know of what each and all of the variables are in PV=nRT? R is faraday's consistent, 0.08206 - continually. additionally, do no longer overlook, in ALL of those varieties of formulae, that temp must be switched over to Kelvin (C+273). stable luck!