Question

In a thermodynamic process, pressure of a fixed
mass of a gas is changed in such a manner that
the gas molecules gives out 20 J of heat and
10 J of work is done on the gas. If the initial
internal energy of the gas was 40 J, then the
final internal energy will be
​

Answer 1

Answer: this is the answer:

dQ = -20J, dW = -10 J,

U 1=40J

By 1st law of thermodynamics,

dQ = dU + dW

⇒dU = dQ-dW

= -20 + 10 = -10J

Final internal energy,

U2

U1

dU

= 40 - 10 = 30 J

hope it helps.....

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Explanation:

Answer 2

Answer:

dQ = -20J, dW = -10 J,

U 1=40J

By 1st law of thermodynamics,

dQ = dU + dW

⇒dU = dQ-dW

= -20 + 10 = -10J

Final internal energy,

U2

U1

dU

= 40 - 10 = 30 J

hope it helps.....

please mark brainliest..