Question

A 94 L sample of neon gas (molar mass= 20.2 g/mol) has a mass of 44 g at a pressure of 150 kPa. What is its temperature in Kelvins?

Answer 1

778.96 K.

Step-by-step 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 = 150.0 kPa/101.325 = 1.48 atm, V = 94.0 L, R = 0.082 L.atm/mol.K, n = mass/molar mass = (44.0 g)/(20.2 g/mol) = 2.178 mol.

∴ T = PV/nR = (1.48 atm)(94.0 L)/(2.178 mol)(0.082 L.atm/mol.K) = 778.96 K.