Animated Solution for Physics - Thermodynamics: List-I describes thermodynamic processes in four different systems. List-II gives the magnitudes (either exactly or as a close approximation) of possible changes in the internal energy of the system due to the process.

Thermodynamics is the study of energy in transit. It governs everything from the engines in our cars to the beating of our hearts. In this epic problem, we are tasked with finding the change in internal energy ($\Delta U$) for four completely different physical systems. It is like a grand tour of the First Law of Thermodynamics!

The Boiling Cauldron

Imagine a tiny amount of water, just $10^{-3}\text{ kg}$ (or 1 gram), sitting at its boiling point of $100^{\circ }\text{C}$. We are pumping heat into it to convert it entirely into steam.

First, how much heat are we actually supplying? We use the latent heat of vaporization. The heat supplied is $\Delta Q=mL$. Plugging in the values, we get $\Delta Q=10^{-3}\times 2250\text{ kJ/kg}=2.25\text{ kJ}$.

But wait, the steam takes up much more space than the liquid water! It expands against the atmospheric pressure of $10^5\text{ Pa}$. This means the system is doing work on the universe. The work done is $\Delta W=P\Delta V$. The volume changes from a minuscule $10^{-6}{\text{ m}}^3$ to $10^{-3}{\text{ m}}^3$.

Calculating this, $\Delta W=10^5\times (10^{-3}-10^{-6})\approx 100\text{ J}=0.1\text{ kJ}$.

According to the First Law of Thermodynamics, $\Delta U=\Delta Q-\Delta W$. The internal energy increases by $2.25-0.1=2.15\text{ kJ}$. The closest match is $2\text{ kJ}$.

The Expanding Diatomic Gas

Next, we have $0.2$ moles of a rigid diatomic gas expanding at constant pressure. The volume triples from $V$ to $3V$.

Because the pressure is constant, Charles's Law tells us that the temperature must also triple! The initial temperature is $500\text{ K}$, so the final temperature becomes $1500\text{ K}$. The change in temperature is a massive $\Delta T=1000\text{ K}$.

For a rigid diatomic gas, the molar heat capacity at constant volume is $C_V=\frac{5}{2}R$.

Even though the process is isobaric, the change in internal energy for an ideal gas is always given by $\Delta U=n{C}_V\Delta T$.

Substituting our values: $\Delta U=0.2\times \left(\frac{5}{2}\times 8.0\right)\times 1000=4000\text{ J}=4\text{ kJ}$.

The Swift Adiabatic Crush

Now, imagine a monatomic gas being compressed so rapidly that no heat can escape. This is an adiabatic process ($\Delta Q=0$).

The volume is crushed from $V$ to $V/8$. To find the new pressure, we use the adiabatic relation $P_1{V}_1^{\gamma }=P_2{V}_2^{\gamma }$. For a monatomic gas, $\gamma =\frac{5}{3}$.

$P_2=P_1{\left(\frac{V_1}{V_2}\right)}^{\gamma }=2\text{ kPa}\times (8{)}^{5/3}$. Since $8$ is $2^3$, $8^{5/3}$ is simply $2^5=32$. So, the final pressure shoots up to $64\text{ kPa}$!

The change in internal energy is exactly the negative of the work done: $\Delta U=\frac{P_2{V}_2-P_1{V}_1}{\gamma -1}$.

Plugging in the pressures and volumes, we get $\Delta U=\frac{3}{2}\left(64\times \frac{1}{24}-2\times \frac{1}{3}\right)=\frac{3}{2}\left(\frac{8}{3}-\frac{2}{3}\right)=3\text{ kJ}$.

The Vibrating Molecules

Finally, we encounter a diatomic gas that isn't rigid. Its molecules are vibrating! This adds 2 extra degrees of freedom, bringing the total to $f=7$.

This means $C_V=\frac{7}{2}R$ and $C_P=\frac{9}{2}R$.

We are told that $9\text{ kJ}$ of heat is supplied at constant pressure. So, $\Delta Q=n{C}_P\Delta T=9\text{ kJ}$.

We need $\Delta U$, which is $n{C}_V\Delta T$. Notice the beautiful shortcut here: the ratio $\frac{\Delta U}{\Delta Q}$ is simply $\frac{C_V}{C_P}=\frac{7}{9}$.

Therefore, $\Delta U=\frac{7}{9}\times 9\text{ kJ}=7\text{ kJ}$.

We have successfully navigated four distinct thermodynamic landscapes. By trusting the First Law and understanding the degrees of freedom, we matched every process perfectly. This is the true power of physics!