Question

In a thermal process 400J of heat is given to a gas and 100J of work is also done on it .Change in internal energy of the system is?​

Answer 1

Answer:

Hola✌️♥️☺️

Explanation:

✏️dQ=du+dwdQ=du+dw By sign convention

dQ=+400JquaddQ=+400Jquad Heat absorbed is

Positive

dw=−100Jdw=−100J heat given out is

Negative

du=dQ−dwdu=dQ−dw Work done by the system is positive

=400−(−100)=400−(−100) Work done on the system is negative

=500J

✒️Hope it works☺️♥️✌️...

Answer 2

Answer:

heya dear...☺️✌

Explanation:

hope dis helps...btw telugu?

have a great dayh....