Question

A closed system receives an input heat of 450kj and increases the internal energy of the system for 325kj ?​

Answer 1

Step-by-step explanation:





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Question



A closed system receives 200 kJ of constant volume. It then rejects 100 kJ of heat while it has 50 kJ of work done on it at constant pressure. If an adiabatic process can be found which can be found which will restore the system to its initial state, the work done by the system during this process is

A

100 kJ

B

50 kJ

C

150 kJ

Correct Answer

D

200 kJ

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Solution



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Correct option is

C

150 kJ

Through the first process,

200 kJ at constant volume.

Hence internal energy increases by 200 kJ, 

For second process:

q=−100 kJ

w=−50 kJ

Hence ΔU=−50kJ

Including both step one and two, change in IE is 150 kJ

Now to go back to initial state, the Internal energy must decrease by 150 kJ

Given that process is adiabatic.

Using 1st law; q=ΔU+w

0=−150+w

w=150 kJ