Liquid chlorobenzene goes through a process where there is negligible heat flow and no work done on the surroundings. initially it is at 330 k and 3,500 kpa. at the end of the process the final pressure is 2,000kpa. estimate the temperature change and the entropy change of chlorobenzene.

The change in temperature is -141.428K

The entropy change is -60.75J/mole K Kg

Explanation:

Given that,

Initial Temperature, T₁ = 330k

Initial Pressure, P₁ = 3500kpa

Final pressure, P₂ = 2000kpa

Since the work done on the surrounding is zero

hence, the change in volume of the system will be zero and the process will be constant

Calculate the Temperature of the system

Substitute the value in the above equation

Calculate the change in temperature

ΔT = T₂ - T₁

     = 188.57K - 330K

      = -141.428K

The change in temperature is -141.428K

Expression of change in entropy

ds is the entropy change

equation for

Substitute for dQ in the expression of change in entropy and integrate

substitute the values

The entropy change is -60.75J/mole K Kg

okay i have so many questions. but the big one here is: um wt..f is this mess? ?