Determine the temperature of 1.42 mol of a gas contained in a 3.00-L container at a pressure of 123 kPa.

31.2K

Explanation:

The following data were obtained from the question:

n = 1.42 mol

V = 3L

P = 123 kPa = 123000Pa

Recall: 101325Pa = 1atm

123000Pa = 123000/101325 = 1.21atm

R = 0.082atm.L/Kmol

T =?

Using the ideal gas equation PV = nRT, we can obtain the temperature as follows:

PV = nRT

T = PV/nR

T = (1.21 x 3)/(1.42 x 0.082)

T = 31.2K

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