Question

If the change in internal energy of an ideal gas is equal to negative of the work done by the system, then
A. the process may be adiabatic
B. the process may be isothermal
C. the process must be isobaric
D. the temperature must decrease

Answer 1

Hint: We know from the first law of thermodynamics that the change in the internal energy of a closed system is equal to the summation of the heat added to the system and work done. This can be represented as $\Delta U = Q + W$ where $\Delta U$ is the change in the internal energy, $Q$ is net heat if the heat is added to the system it will have positive sign and if heat leaves the system it will have negative sign, $W$ is work done if work is done by the system it will have negative sign and if the work done on a system it will have positive sign.

Complete step by step answer:
We have studied that in adiabatic processes there is no transfer of heat and mass between the surrounding and system. In an isothermal process the temperature of the system remains constant whereas in the isobaric process pressure remains constant which means $\Delta P = 0$. In the problem it is given that the change in internal energy of an ideal gas is equal to negative of the work done by the system which means $\Delta U = - W$ hence here $Q = 0$. Now we know that in adiabatic processes the net heat used to be zero so the process may be adiabatic.

So, the correct answer is “Option A”.

Note: We have solved this problem with the help of the first law of thermodynamics which gives a relationship among internal energy, net heat and work done. We also know about the types of the adiabatic processes so the adiabatic process seems to be most suitable here because it is given that the change in internal energy of an ideal gas is equal to negative of the work done by the system.