Question

9.) What is the density of Ar gas, in units of g/L, at a pressure of 662 mm Hg and a temperature of 25.0°C ?

Answer 1

The pressure of Ar gas = 662 mmHg x ( 1 atm /760 mmHg) = 0.871 atm

Temperature(T) = 25.0 ^oC + 273.15 = 298.15 K

The ideal gas law equation is as follows:

PV = nRT ----------(1)

Here, Pressure is P, Volume is V, Number of moles is n, Gas constant is R and Temperature is T.

The formula to calculate number of moles is as follows:

Number of moles(n) = Mass(m) / Molar mass(M)

Substitute n = (m/M) in equation (1) as follows:

PV = (m/M)RT

PM =(m/V)RT

PM = dRT ----------Since, Density(d) = Mass(m) / Volume(V)

Rearrange the formula for density as follows:

d = PM /RT

Substitute 0.871 atm for P, 39.948 g/mol for M, 0.08206 L.atm/mol.K for R and 298.15 K for T and determine the density of the gas as follows:

d =(0.871 atm x 39.948 g/mol) /(0.08206 L.atm/mol.K x 298.15 K)

d = 1.42 g/L