Question: (II) In an engine, an almost ideal gas is compressed adiabatically to half its volume. In doing so, 2630 J of work is done on the gas. (a) How much heat flows into or out of the gas? (b) What is the change in internal energy of the gas? (c) Does its temperature rise or fall?

According to the first law of thermodynamics, the change in the closed system’s internal energy can be expressed as:

\(\Delta U = Q - W\)

Here,\(Q\)is the heat added and\(W\)is the work done.

Given data:

The work done on the gas is \(W = - 2630\;{\rm{J}}\).

(a)

There is no heat exchange happens between the surroundings and system in the adiabatic compression process. Hence, the heat that flows into or out of the gas is zero. That is \(Q = 0\).

(b)

The change in the internal energy of the gas can be calculated as:

\(\begin{aligned}{c}\Delta U &= Q - W\\\Delta U &= 0 - \left( { - 2630\;{\rm{J}}} \right)\\\Delta U &= 2630\;{\rm{J}}\end{aligned}\)

Thus, the change in the internal energy of the gas is \(2630\;{\rm{J}}\).

(c)

As the gas is compressed adiabatically, there will be an increment in the temperature of the gas. Moreover, from part (b), it is clear that the change in internal energy is positive, which indicates a rise in internal energy. Temperature varies directly with internal energy. Therefore, as internal energy rises, the temperature also rises.

Hence, the temperature rises during the adiabatic compression process.