Question

The temperature of one mole of an ideal gas increases from 298K to 308K when it absorbs 200J of heat at constant volume. The change in the internal energy of the gas is:
A) 200 J
B) 140 J
C) -200 J
D) -140 J

Answer 1

Hint:
The internal energy is the measure of energy within a thermodynamic system, excluding its kinetic and potential energy. But it includes any gain or loss of energy due to any change in its internal state.

Formula used
Work done:
                                                              $W = - P\Delta V\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(1)$
The first law of thermodynamics states that the change in internal energy of a system equals the net heat transfer into the system plus the net work done on the system.
Change in internal energy:
                                                              $\Delta U = q + W\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,(2)$

Complete step by step answer:
The first law of thermodynamics is based on the law of conservation of energy. According to it the total energy of a system always remains conserved, it can neither be created nor destroyed but it can be transformed from one form to another.
The First Law states that the internal energy of a system has to be equal to the work that is being done on the system, plus or minus the heat that flows in or out of the system and any other work that is done on the system.
Mathematically it is written as: $\Delta U = q + W$, where q and W vary depending on how the change is being carried out.
The question asks us to find the change in internal energy and for that we need the amount of heat produced and work done.
We can find the amount of work done from equation (1): $W = - P\Delta V$
According to the question the volume is constant so the change in volume will be 0.
So, W = -P(0) =0
Hence the amount of work done is zero.
The question has already given the value of heat produced, which is = 200 J.
Now, put the values of work done and heat in equation (2):
                                                   $\Delta U = q + W$
    Since W = 0, $\Delta U = q$
                                                                      = 200 J
The change in internal energy is 200 J.

So, the correct option is: (A).

Note:
Always be careful while putting the values and their signs in the equations. The value of ‘q’ is negative when the system releases heat and it is positive when the system absorbs heat. Also W is negative when work is done by the system and it is positive when work is done on the system.