Part 1. Determine the molar mass of a 0.314-gram sample of gas having a volume of 1.6 L at 287 K and 0.92 atm. Show your work. Part 2. Describe the temperature and pressure conditions at which the gas behaves like an ideal gas.

Part 1: 1.721 g/mol.

Part 2: at high T and low P.

Explanation:

Part 1:

We can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm (P = 0.92 atm).

V is the volume of the gas in L (V = 1.6 L).

n is the no. of moles of the gas in mol (n = ??? mol).

R is the general gas constant (R = 0.0821 L.atm/mol.K),

T is the temperature of the gas in K (T = 287 K).

∴ n = PV/RT = (0.92 atm)(1.6 L)/(0.0821 L.atm/mol.K)(287 K) = 0.1825 mol.

To find the molar mass, we can use the relation:

n = mass/molar mass

∴ molar mass of the gas = mass/n = (0.314 g)/(0.1825 mol) = 1.721 g/mol.

Part 2:

The gas is considered to be ideal gas when:

1. Its particles have zero volume.

2. have intermolecular forces are also zero.

But, really the gas has significant volume (volume of the container decreases) and intermolecular forces (pressure of the gas increases), so it deviates from ideal behavior. If the real gas is low pressure and reasonably high temperature then it will behave like an ideal gas in that our measuring equipment will not be accurate enough to measure a difference. As the pressure gets higher or the temperature gets low enough, the differences between an ideal gas and a real gas become measurable.