Animated Solution for Chemistry - Chemical Thermodynamics: Five moles of an ideal gas at $293\text{ K}$ is expanded isothermally from an initial pressure of $2.1\text{ MPa}$ to $1.3\text{ MPa}$ against at constant external pressure $4.3\text{ MPa}$. The heat transferred in this process is ......... ${\text{kJ mol}}^{-1}$. (Rounded off to the nearest integer) $[R=8.314{\text{ J mol}}^{-1}{\text{K}}^{-1}]$

The problem of isothermal expansion against a constant external pressure is a classic application of the First Law of Thermodynamics. Let's break down the physical reality of what's happening inside the cylinder and how the math perfectly models it.

Imagine a sturdy cylinder fitted with a movable piston. Inside, we have 5 moles of an ideal gas. The entire system is maintained at a constant temperature of 293 K. This is our first major clue: the process is isothermal.

The gas is initially at a high pressure of $2.1\text{ MPa}$. It pushes against the piston, expanding until its internal pressure drops to $1.3\text{ MPa}$. However, the expansion isn't happening in a vacuum or against a slowly changing force. The gas is pushing against a heavy, constant external pressure of $4.3\text{ MPa}$. Because the external pressure is constant and significantly different from the internal pressure throughout the process, this expansion is highly irreversible.

To find the heat transferred ($Q$), we must invoke the First Law of Thermodynamics:

For an ideal gas, the internal energy ($U$) is solely a function of its temperature. Since our process is isothermal, the temperature doesn't change, which means the internal energy doesn't change either ($\Delta U=0$).

This simplifies our master equation beautifully:

This tells us a profound physical truth: all the work done by the gas during the expansion is exactly compensated by the heat flowing into the gas from the surroundings to keep its temperature constant.

The work done by a gas expanding against a constant external pressure is given by:

We don't know the initial and final volumes ($V_1$ and $V_2$), but we do know the pressures. We can use the Ideal Gas Law ($pV=nRT$) to express volume in terms of pressure:

Substituting this into our work equation gives us a powerful new form:

Now, we substitute our known values. The external pressure $p_{\text{ext}}$ is $4.3\text{ MPa}$, $n=5\text{ mol}$, $R=8.314{\text{ J mol}}^{-1}{\text{K}}^{-1}$, and $T=293\text{ K}$. The initial and final pressures are $2.1\text{ MPa}$ and $1.3\text{ MPa}$ respectively.

Notice a beautiful mathematical convenience here: because $p_{\text{ext}}$ is in the numerator and $p_1,p_2$ are in the denominator, their units of Mega Pascals (MPa) will perfectly cancel out! We don't need to convert them to Pascals.

Let's resolve the bracket first:

Plugging this back in:

Converting this to kilo Joules, we get $W\approx -15.35\text{ kJ}$.

Since $Q=-W$, the heat transferred into the system is $15.35\text{ kJ}$.

The question asks for the heat transferred in ${\text{kJ mol}}^{-1}$. Strictly speaking, we should divide our total heat by the 5 moles of gas, which would give us $3.07{\text{ kJ mol}}^{-1}$.

However, in the official JEE Main answer key, the total heat transferred ($15\text{ kJ}$) was accepted as the correct integer answer. This implies the unit ${\text{kJ mol}}^{-1}$ in the question text was likely a typographical error intended to just be $\text{kJ}$. Therefore, rounding $15.35$ to the nearest integer, we arrive at our final answer of 15.