One mole of an ideal gas expands adiabatically from an initial state (TA, V0) to final state (Tf, 5V

Q: Why is the final temperature $T_f$ equal to the initial temperature $T_B$ for the second process?

The problem states that the second mole of gas expands *isothermally*. By definition, an isothermal process occurs at a constant temperature. Therefore, the temperature at the beginning of the process ($T_B$) must be exactly the same as the temperature at the end ($T_f$).

Q: Why do we use the equation $TV^{\gamma-1} = \text{constant}$ instead of $PV^\gamma = \text{constant}$ for the adiabatic process?

While both equations are valid for a reversible adiabatic process of an ideal gas, the question specifically asks for a ratio of *temperatures* ($T_A/T_B$) and provides information about the initial and final *volumes* ($V_0$ and $5V_0$). Using the $T-V$ relation directly saves us the extra step of calculating pressures using the ideal gas law.

Q: On a P-V diagram, why does the adiabatic curve start at a higher pressure than the isothermal curve if they end at the same point?

An adiabatic expansion is "steeper" than an isothermal expansion on a P-V diagram. Because no heat is added during an adiabatic expansion, the gas cools down, causing the pressure to drop more rapidly than it would in an isothermal process where heat is absorbed to maintain the temperature. To end up at the exact same final pressure and volume, the adiabatic process must have started from a higher initial pressure.

Q: Does the fact that we have "one mole" of gas in each process affect the final ratio?

In this specific problem, the number of moles ($n=1$) doesn't explicitly change the algebraic steps because the $TV^{\gamma-1} = \text{constant}$ relation is independent of $n$. However, it ensures that the heat capacities and the ideal gas constant behave consistently across both samples of the gas.

Q: What if the gas was specified as monatomic or diatomic? How would that change the answer?

The final algebraic expression $5^{\gamma-1}$ would remain the same. However, you would be able to calculate a numerical value. For a monatomic gas, $\gamma = 5/3$, so the ratio would be $5^{2/3}$. For a diatomic gas, $\gamma = 7/5$, making the ratio $5^{2/5}$. The beauty of this question is that it generalizes the result for any ideal gas!

Question

One mole of an ideal gas expands adiabatically from an initial state $(T_A, V_0)$ to final state $(T_f, 5V_0)$. Another mole of the same gas expands isothermally from a different initial state $(T_B, V_0)$ to the same final state $(T_f, 5V_0)$. The ratio of the specific heats at constant pressure and constant volume of this ideal gas is $\gamma$. What is the ratio $T_A / T_B$?

Options:

(A) $5^{\gamma-1}$

(B) $5^{1-\gamma}$

(C) $5^\gamma$

(D) $5^{1+\gamma}$

Sigma Percentile · Accepted Answer

The beauty of thermodynamics lies in its ability to connect abstract mathematical equations to physical realities. Imagine you are standing in a laboratory, observing two identical samples of an ideal gas. Both samples are destined to reach the exact same final state, but they will take entirely different thermodynamic paths to get there.

This problem is a classic test of your conceptual clarity regarding thermodynamic processes. It doesn't require pages of complex calculus; instead, it demands a sharp understanding of the equations of state for isothermal and adiabatic expansions. Let's embark on this journey and decode the physics step by step.

### Analyzing the Setup

We are given two distinct processes occurring on one mole of an ideal gas.

**Process 1:** An adiabatic expansion from an initial state $(T_A, V_0)$ to a final state $(T_f, 5V_0)$.
**Process 2:** An isothermal expansion from a different initial state $(T_B, V_0)$ to the exact same final state $(T_f, 5V_0)$.

Our ultimate goal is to find the ratio of the initial temperatures, $\frac{T_A}{T_B}$. To do this, we need to extract mathematical relationships from the physical constraints of each process.

### The Isothermal Journey

Let's start with the simpler of the two: the isothermal process. The word "isothermal" literally translates to "constant temperature." When a gas expands isothermally, it absorbs heat from its surroundings at a rate that perfectly balances the work it does, ensuring its internal energy—and thus its temperature—remains completely unchanged.

Mathematically, this means the initial temperature must equal the final temperature. For our second mole of gas, the initial temperature is $T_B$ and the final temperature is $T_f$. Therefore, we can immediately write our first crucial equation:

$$T_B = T_f$$

This simple equality is the linchpin of the entire problem. It gives us a direct bridge between the two processes.

### The Adiabatic Plunge

Now, let's turn our attention to the adiabatic process. In an adiabatic expansion, the gas is perfectly insulated. It does work on its surroundings, but no heat is allowed to enter or leave the system. As a result, the gas must expend its own internal energy to do this work, causing its temperature to drop rapidly.

For a reversible adiabatic process involving an ideal gas, the relationship between temperature and volume is governed by Poisson's equation:

$$T V^{\gamma-1} = 	ext{constant}$$

Here, $\gamma$ (gamma) is the ratio of specific heats ($\frac{C_p}{C_v}$). This equation tells us that as the volume $V$ increases, the temperature $T$ must decrease to keep the product constant.

Let's apply this master equation to the initial and final states of our first mole of gas. The initial state is $(T_A, V_0)$ and the final state is $(T_f, 5V_0)$. Substituting these into our adiabatic relation yields:

$$T_A V_0^{\gamma-1} = T_f (5V_0)^{\gamma-1}$$

### The Grand Synthesis

We now have two powerful equations. The adiabatic equation contains $T_f$, but we want our final answer in terms of $T_B$. This is where our isothermal insight comes into play.

Since we established earlier that $T_f = T_B$, we can seamlessly substitute $T_B$ into our adiabatic equation. Let's make the swap:

$$T_A V_0^{\gamma-1} = T_B (5V_0)^{\gamma-1}$$

Suddenly, the equation only contains the variables we care about: $T_A$, $T_B$, $V_0$, and $\gamma$. The physics is complete; all that remains is the final mathematical stroke.

### Final Calculation

Our objective is to isolate the ratio $\frac{T_A}{T_B}$. Let's rearrange the equation by dividing both sides by $T_B$ and by $V_0^{\gamma-1}$:

$$\frac{T_A}{T_B} = \frac{(5V_0)^{\gamma-1}}{V_0^{\gamma-1}}$$

Using the properties of exponents, we can group the volume terms together:

$$\frac{T_A}{T_B} = \left(\frac{5V_0}{V_0}ight)^{\gamma-1}$$

The initial volume $V_0$ elegantly cancels out from the numerator and the denominator, leaving us with a pure, dimensionless ratio:

$$\frac{T_A}{T_B} = 5^{\gamma-1}$$

And there we have it! The ratio of the initial temperatures is simply $5^{\gamma-1}$. This result is profound because it shows that the temperature ratio depends entirely on the expansion ratio (which is 5) and the nature of the gas (dictated by $\gamma$). Whether the gas is monatomic, diatomic, or polyatomic, this elegant formula holds true.

Select Answer:

Visualized Solution

$P - V Diagram$

$T = constant$

$T_{B} = T_{f}$

$T V^{γ - 1} = constant$

$T_{A} V_{0}^{γ - 1} = T_{f} (5 V_{0})^{γ - 1}$

$T_{A} V_{0}^{γ - 1} = T_{B} (5 V_{0})^{γ - 1}$

$\frac{T _{A}}{T _{B}} = 5^{γ - 1}$

$Conclusion$

The Sigma Insight: Thermodynamic Processes

Analyzing the Setup

The Isothermal Journey

The Adiabatic Plunge

The Grand Synthesis

Final Calculation

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Select Answer:

Visualized Solution

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } P−V Diagram

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } T=constant

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } TB​=Tf​

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } TVγ−1=constant

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } TA​V0γ−1​=Tf​(5V0​)γ−1

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } TA​V0γ−1​=TB​(5V0​)γ−1

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } TB​TA​​=5γ−1

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Conclusion