Answer to Question #234540 in Molecular Physics | Thermodynamics for Unknown346307

Question

.14. A closed system of constant volume experiences a temperature rise of 25°C

when a certain process occurs. The heat transferred in the process is 30 kJ. The specific heat at

constant volume for the pure substance comprising the system is 1.2 kJ/kg°C, and the system

contains 2.5 kg of this substance. Determine :

(i) The change in internal energy ;

(ii) The work done.

Answer 1

Determine the internal energy:

$ΔU=mc_v \times ΔT \\ = 2.5 \times 1.2 \times 25 \\ = 75 \;kJ$

Determine the work done:

$Q= W + ΔU \\ 30= W + 75 \\ W= -45 \;kJ$

Probably the work input is given to the system as follows:

Determine the work done using PdV relation:

$W = PdV \\ = P \times 0 \\ = 0 \; kJ$

The above relation cannot be used here becaunse, 30 kJ heat addition causes an increase in internal energy of 75 kJ. The violates the law of energy conservation. Hence, the remaining 45 kJ of energy must be supplied to the system using through work.

Therefor, the work done is W = - 45 kJ.

Note: If we get two answers for the same variable then take the answer which obeys the fundamental laws like:

(a) Law of energy.

(b) Law of mass conservation.