Question

At what temperature will 0.0100 mole of argon gas have a volume of 275 ml at 100.0 kpa?

Answer 1

Explanation:

For an ideal gas, product of pressure and volume equals n times R times T.

Mathematically, PV = nRT

where P = pressure

V = volume

n = number of moles

R = gas constant

T = temperature

Also, it is known that in 1 kpa there are 0.0098 atm. So, 100.0 kpa equals 0.98 atm. And 275 ml equals 0.275 l. Therefore, calculate temperature as follows.

PV = nRT

`0.98 * 0.275 l = 0.01000 mol * 0.082 L atm mol^(-1)K^(-1) * T`

T =
`32.86 K`

Thus, we can conclude that temperature for the given problem is
`32.86K`.

Answer 2

the main formula is
PxV = nx RxT
P: pressure=100.0 Kpa, and 1 atm = 101.325 kilopascal (kPa), so 100Kpa=0.9atm
V :volume=0.275l
n: number of moles= 0.0100mole
R: Gas constant=8.31
T: temperature

so T = PxV /nx R=0.9 x0.275 / 0.01 x8.31=2.97°, so T=2.97°+273=275.97K