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

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.

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;

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.

here we have to get the change of moles on decrease the volume of the reaction in gas phase.

the   per the given options it will decrease by (d) 11-50%

we know, on decreasing the volume of the gas phase reaction the reaction moves to the product side.

the reaction:                     so₂cl₂ (g) = so₂ (g) + cl₂ (g)         total

the moles at equilibrium:   1 - α                 α               α           = 1+α

concentrations:                   1 - α               α               α           = 1 + α

(concentration = moles/volume, as volume is 1 l)

(where α is the degree of dissociation)

now the dissociation constant in terms of concentration is :

=  

or,

or,   = (at v = 1l)

the reaction:                     so₂cl₂ (g) = so₂ (g) + cl₂ (g)         total

the moles at equilibrium:   1 - α                 α               α           = 1+α

concentrations:               (1 - α)/0.5           α/0.5         α/0.5       = (1 + α)/0.5

thus, the dissociation constant in terms of concentration is :

or,  

=  

so, the change of 1 : 1/0.5

or, 1: 2

thus it will decrease by 50%.

or, as per the given options it will decrease by 11-50%

if the concentration of products is decreased, the equilibrium will shift toward the products). add an inert gas (one that is not involved in the reaction) to the constant-volume reaction mixture: this will increase the total pressure of the system, but will have no effect on the equilibrium condition.

explanation: