A 3.00 L flask contains 2.33 g of argon gas at 312 mm Hg What is the temperature of the gas

T = 257.36 k

Explanation:

using the ideal gas law

pv=nRt

first, convert pressure from mmhg to kpa

312 x (101.3/760) = 41.58 kpa

R is constant = 8.31

to get n (number of moles)

n=m/M (m is mass, M is molar mass)

molar mass of argon is 39.948

n = 2.33/39.948

n=0.0583

substitute;

41.58 x 3 = 0.0583 x 8.31 x T

T = (41.58 x 3) / (0.0583 x 8.31)

T = 257.36 k