Question

On a hot summer day, the density of air at atmospheric pressure at 35.5°C is 1.1970 kg/m3. (a) What is the number of moles contained in 1.00 m3 of an ideal gas at this temperature and

Answer 1

Complete question:

On a hot summer day, the density of air at atmospheric pressure at 35.5°C is 1.1970 kg/m3. (a) What is the number of moles contained in 1.00 m3 of an ideal gas at this temperature and pressure.

Answer:

The  number of moles contained by an ideal gas at this temperature and pressure is 41.32 moles.

Explanation:

Given;

density of dry air, ρ = 1.1970 kg/m³

temperature of the air, T = 35.5°C  = 273 + 35.5 = 308.5 K

air volume, V = 1 m³

Apply ideal gas law for dry to calculate the air pressure;

$P = \rho R_dT$

where;

P is the air pressure

ρ is the air density

Rd is gas constant for dry air = 287 J/kg/K

P = 1.197 x 287 x 308.5 = 105,981.78 Pa

(a) Now, determine the number of moles contained by an ideal gas at this temperature and pressure, by applying ideal gas law;

PV = nRT

where;

P is the pressure of the gas (Pa)

V is the volume of the gas (m³)

n is number of gas moles

R is gas constant = 8.314 m³.Pa / mol.K

T is temperature (K)

n = (PV) / (RT)

n = (105,981.78 x 1) / (8.314 x 308.5)

n = 41.32 moles

Therefore, the  number of moles contained by an ideal gas at this temperature and pressure is 41.32 moles.

Answer 2

The number of moles of an ideal gas at this temperature and pressure is 41.5 moles.

Given that;

Density of dry air = 1.1970 kg/m3

Pressure of dry air = ?

Temperature of dry air = 35.5°C + 273 = 308.5 K

Hence;

P = Density × gas constant of dry air × Temperature

P = 1.1970 kg/m3 × 287.1 J/Kg/K × 308.5 K

P = 106019 Pa or 1.05 atm

Using the ideal gas equation;

PV = nRT

n = PV/RT

n = 1.05 atm × 1000 L/0.082 atmL/K.mol × 308.5 K

n = 41.5 moles

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