Air was put into an automobile tire when the tire was cold, at 5.00 C. The tire's air pressure was 3.04 atm. Later, the weather warms up. Also, the automobile is driven, further warming the tire. What will the tire's pressure be when the temperature is 25.0 C?

Explanation:

Let suppose that air inside the tire behaves ideally. The equation of state for ideal gases is:

As tire can be modelled as a closed and rigid container, there are no changes in volume and number of moles. Hence, the following relationship is constructed:

The final pressure is:

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%

the volume of fluid displaced would equal the volume of the gas. use a gas syringe - this is a laboratory instrument that can be used to withdraw a certain volume of gas from a closed system or to measure the volume of gas from a chemical reaction.