Animated Solution for Chemistry - Chemical Thermodynamics: One mole of an ideal monoatomic gas undergoes two reversible processes (A $\to$ B and B $\to$ C) as shown in the given figure : A $\to$ B is an adiabatic process. If the total heat absorbed in the entire process (A $\to$ B and B $\to$ C) is $R{T}_2\ln 10$, the value of $2\log V_3$ is _________. [Use, molar heat capacity of the gas at constant pressure, $C_{p,m}=\frac{5}{2}R$]

Welcome, future engineers and scientists! Today, we are going to dissect a fascinating problem from JEE Advanced 2023. Thermodynamics is not just about memorizing formulas; it's about understanding the story a gas tells as it undergoes various transformations. This problem is a beautiful blend of graph interpretation, the first law of thermodynamics, and a little bit of logarithmic algebra. Let's dive in!

Let's look at the graph provided in the question. It's a Temperature-Volume (T-V) graph. Many students panic when they don't see a standard P-V graph, but T-V graphs are incredibly revealing.

The gas expands from point A to B, and the problem explicitly states this is an adiabatic process. What does that mean physically? It means the gas is perfectly insulated; there is absolutely no heat exchange with the surroundings.

Then, from point B to C, the graph shows a perfectly horizontal line. The temperature remains constant at $60\text{ K}$. A constant temperature process is, by definition, an isothermal process.

To find the volume at state B (which we'll call $V_2$), we need our trusty adiabatic relation connecting temperature and volume:

But wait, we need the value of $\gamma$ (the adiabatic index). The problem gives us the molar heat capacity at constant pressure, $C_{p,m}=\frac{5}{2}R$. This specific value is the signature of a monoatomic gas.

Using the relation $C_p-C_v=R$, we can easily find the molar heat capacity at constant volume:

Now, we can calculate $\gamma$:

Now that we have $\gamma$, let's apply the adiabatic equation between states A and B:

Substitute the known values from the graph ($T_1=600\text{ K}$, $V_1=10{\text{ m}}^3$, $T_2=60\text{ K}$):

Simplify the exponents:

Divide both sides by 60:

Combine the powers of 10 on the left side ($1+\frac{2}{3}=\frac{5}{3}$):

To isolate $V_2$, raise both sides to the power of $\frac{3}{2}$:

Now, let's shift our focus to the second part of the question. We are given the total heat absorbed in the entire process, $q_{\text{net}}=R{T}_2\ln 10$.

According to the principle of superposition, the total heat is the sum of the heat exchanged in each individual process:

Since process $A\to B$ is adiabatic, its heat exchange is exactly zero ($q_{AB}=0$). This is a crucial conceptual step! It means all the heat absorbed comes entirely from the isothermal process $B\to C$.

For a reversible isothermal process, the heat absorbed is equal to the work done by the gas, which is given by the formula:

We know we have $1$ mole of gas ($n=1$), and the temperature is $T_2$. Let's equate this to the given total heat:

This equation simplifies beautifully. The $R{T}_2$ terms cancel out on both sides:

For the natural logarithms to be equal, their arguments must be equal:

Now, substitute the value of $V_2$ we found earlier:

We are almost at the finish line! The question asks for the value of $2\log V_3$. Here is a classic trap: notice the base! The problem uses $\ln$ for natural logarithm (base $e$) and $\log$ for common logarithm (base 10).

Substitute $V_3$ into the expression:

Using the power rule of logarithms, bring the exponent to the front:

Since ${\log }_{10}10=1$, the twos cancel out, leaving us with: Final Answer = 7

This was a brilliant problem that seamlessly combined graph interpretation with the first law of thermodynamics. Keep practicing, stay curious, and always watch out for those logarithmic bases!