Question: An ideal gas undergoes an isothermal expansion from state A to state B. In this process (use sign conventions, page 413),

(a) \[Q = 0,\;\Delta U = 0,\;W > 0\].

(b) \[Q > 0,\;\Delta U = 0,\;W < 0\].

(c) \[Q = 0,\;\Delta U > 0,\;W > 0\].

(d) \[Q > 0,\;\Delta U = 0,W > 0\].

(e) \[Q = 0,\;\Delta U < 0,\;W < 0\].

The internal energy of the system is the addition of the work done and the heat added to the system. If the value of the heat added to the system and work done is more, then the value of the internal energy is more.

The expression for the internal energy of a monoatomic ideal gas is as follows:

\[\Delta U = \frac{3}{2}nR\left( {\Delta T} \right)\]

Here, \(n\)is the number of moles,\(R\) is the universal gas constant, and\[\Delta T\]is the difference in the temperature.

In the isothermal process, the temperature remains constant. Therefore, \[\Delta T = 0\]. Hence, the value of the change in the internal energy will be \[\Delta U = 0\].

The expression for the internal energy according to first law of thermodynamics is as follows:

\[\Delta U = Q - W\]

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

Since,\[\Delta U = 0\]. So, the above equation can be written as:

\[Q = W\]

In the process of isothermal expansion, the gas does work on surroundings. Thus, the work done is positive. This implies that\[W > 0\]. Since,\[Q = W\], the value of the heat added also will be positive, and so\[Q > 0\].

Thus, option (d) is the correct answer.