Question

In adiabatic process, the work done by system is 50J, then:
(a) The temperature of the system will increase by 50J.
(b) The temperature of the system will remain constant.
(c) The internal energy of the system will increase by 50J.
(d) The internal energy of the system will decrease by 50J.

Answer 1

Apply 1st law of thermodynamics. Remember in adiabatic process, heat and mass exchange between the system and surrounding is 0.
Formula used:
1st law of thermodynamics:
$dQ = dU + dW$ …… (1)
where,
dQ is the heat supplied.
dU is the change in internal energy of the system.
dW is the work done by the system.

Internal energy:
$dU = {C_V}dT$ …… (2)
where,
$C{}_V$ is the specific heat at constant value.
dT is the change in temperature of the system.

Complete step-by-step answer:
Given:
1. Work done by the system (dW) = +50J
(Work done is taken to be positive as the sign convention in physics is: Work done by the system is taken as positive)

To find: The temperature change and change in internal energy of the system.

Step 1 of 3:
In adiabatic process, heat supplied is 0:
$dQ = 0$ ……(3)
Use eq (1) and substitute eq (3):
$dU = - dW$ ……(4)

Step 2 of 3:
Substitute the value of dW in eq (4):
$dU = - 50J$
Negative dU implies that internal energy has decreased. Thus, internal energy decreased by 50J.

Step 3 of 3:
Use eq (2) to find the sign of dT:
$- 50 = {C_V}dT$
We know that, $C{}_V$ is a positive quantity. Therefore, dT will be negative. This implies that the temperature will decrease.

Correct Answer:
(d) The internal energy of the system will decrease by 50J.

Note: The 1st law of thermodynamics is about energy conservation in thermodynamic processes in the system and surrounding interaction picture.
- Remember: Work done by the system will be taken as positive and work done on the system will be taken as negative.