Question

à gas receives an amount of heat equal to 110joules and performs 40 joules of work. the change in the internal energy of the gas is

Answer 1

Answer:

70J

Explanation:

dQ = dU+dW

dU = dQ-dW

dU = 110-40

dU = 70J

Answer 2

Given:

• Gas receives 110 Joule of heat
• Gas does 40 Joule of work

To find:

• Change in internal energy ?

Calculation:

In these type of questions, always apply 1ST LAW OF THERMODYNAMICS.

• Since heat is added to system, sign of ∆Q will be positive.

• Also, since the work is done by system, sign of W will also be positive.

So, we can say :

$\therefore\Delta Q = \Delta U + W$

$\implies ( + 110)= \Delta U + ( + 40)$

$\implies 110= \Delta U + 40$

$\implies \Delta U = 110 - 40$

$\implies \Delta U =70 \: joule$

So, change in internal energy is 70 Joule.