An ideal gas contained in a piston which is compressed. the gas is insulated so that no heat flows into or out of it. 1) what happens to the temperature of the gas when it is compressed?

Temperature increases

Explanation: As the gas in the container is an ideal gas so it should follow the ideal gas equation, the equation of state.

We know ideal gas equation to be PV=nRT where

   P=pressure

   V=Volume

   T=Temperature

   R=Real gas constant

   n=Number of moles

since the gas is insulated such that no heat goes into or out of the system .

When we compress the ideal gas using a piston, Thermodynamically it means that work is done on the system by the surroundings.

Now as the ideal gas is been compressed so the volume of the gas would decrease and slowly a time will reach when no more gas can be compressed that is there cannot be any further decrease in volume of the gas.

From the equation PV=nRT

Once there is no further compression is possible hence volume becomes constant so pressure of the ideal gas becomes directly proportional to the temperature as n and R are constants. Also as the pressure and volume are inversely related so an decrease in volume would lead to an increase in pressure.

As the ideal  gas is compressed so the pressure of the gas would increase since the gas molecules have smaller volume available after compression hence the gas molecules would quite frequently have collisions with other gas molecules or piston and this collision would lead to increase in speed of the gas molecules and  so the pressure would increase .

The increase in pressure would lead to an increase in temperature as show by the above ideal gas equation because the pressure and temperature are directly related.

So here we can say that work done on the system by surroundings leads to increase in temperature of the system.

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%