If 0.230 moles of a gas occupies a 116.2 Liter container at 456 mm Hg, what's the temperature?

Answer 1

hey:
Using the ideal gas law we can solve for temperature knowing the pressure and moles of the gas!

Ideal Gas Law= PV=nRT
P= pressure (in atm)
V= volume ( i litres)
N= number of moles
R= 0.821 , universal gas constant
T= temperature in kelvins

So we have: n=0.230 moles,V=116.2 litres and P=456 mm Hg. We need to:
(1) Convert Pressure to atm
(2) slove for Temperature

(1) Converting mm Hg to Atm

1 atm = 760 mmHg
we have 456mm Hg
So this is equal to: (1/760*456)
O.6 atm

(2) PV=nRT
P=0.6 atm,v=1116.2 L, n=0.230 moles, R=0.0820574587 L · atm · K-1 · mol-1
(0.6)(116.2)=(0.230)(0.0821)T
T=? C -273.15K

by using 1 k= C + 273.15
Good Luck
Sanam

Answer 2

you need to use the ideal gas law

PV=nRT
(press)(Vol)=(moles)(R)(Temp)

pressure needs to be in PASCALS. 760mmHg = 1.01 x 10^5 Pa
volume must be in CUBIC METERS
R = 8.314 J/K/mol.
temperature will be in units of KELVIN

Convert pressure and volume values and substitute into the equation.

Answer 3

PV =NRT
P==PRESSURE
V=VOLUME
N=NO OF MOLES
R CONSTANT-0.0821
T-TEMPERATURE

T =PV/RT HERE P IS ATMOSPHERE-456/760=0.6

T=0.6*116.2/0.0821*0.23
T=3692k

Answer 4

PV=nRT the place P=a million atm, V=559mL(.559L), R=0.082L*atm/ok*mol and T=273K (1atm)(.559L)=n(.082L*atm/ok*mol)(273K) n= (.559L)/(.082L*atm/ok*mol)(273K) n= .0.5 mols .025mols of Hydrogen gasoline occupy a volume of .559L 25mols of Hydrogen gasoline occupy a volume of 559mL.

Answer 5

you need to use the ideal gas law

PV=nRT
(press)(Vol)=(moles)(.0821)(Temp)

pressure needs to be in atmospheres 1atm=760mmHg
temperature needs to be in kelvin K=C+273.15

plug in and solve for T

Answer 6

regular or high test?