25.0 grams of argon are placed in a 20.00L container and the pressure is determine to be 1.250 atm. what is temperature of the gas?

487.02 K.

Explanation:

To solve these problems, we can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant (R = 0.082 L. atm/mol. K),

T is the temperature of the gas in K.

The temperature of the gas = PV/nR.

P = 1.25 atm, V = 20.0 L, R = 0.082 L. atm/mol. K, n = mass/molar mass = (25.0 g) / (39.948 g/mol) = 0.626 mol.

∴ T = PV/nR = (1.25 atm) (20.0 L) / (0.626 mol) (0.082 L. atm/mol. K) = 487.02 K.