Two rigid boxes containing different ideal gases are placed on a table. Box A contains one mole of n

Q: Why is the work done zero in this problem?

The problem explicitly states that the boxes are 'rigid'. A rigid container cannot expand or contract, meaning its change in volume $\Delta V$ is zero. Since work done by a gas is $W = P\Delta V$, the work done is zero.

Q: Why do we use $C_V$ and not $C_P$?

We use $C_V$ (molar heat capacity at constant volume) because the volume of the rigid boxes remains constant during the entire heat transfer process.

Q: How do I know the degrees of freedom for Nitrogen and Helium?

Helium ($He$) is a noble gas and exists as single atoms (monatomic), so it only has 3 translational degrees of freedom. Nitrogen ($N_2$) is a diatomic molecule, which has 3 translational and 2 rotational degrees of freedom at room temperature, totaling 5.

Q: Why does heat flow from Box B to Box A?

Heat always flows spontaneously from a region of higher temperature to a region of lower temperature. Since $\frac{7}{3}T_0$ (approx $2.33 T_0$) is greater than $T_0$, heat flows from Box B to Box A.

Q: What if the problem didn't say 'ignore the heat capacity of boxes'?

If the boxes had significant heat capacity, they would also absorb or release heat. You would have to add terms like $m_{\text{box}} s_{\text{box}} (T_f - T_{\text{initial}})$ to your energy conservation equation to account for the energy stored in the material of the boxes.

Question

Two rigid boxes containing different ideal gases are placed on a table. Box $A$ contains one mole of nitrogen at temperature $T_0$, while box $B$ contains one mole of helium at temperature $(7/3)T_0$. The boxes are then put into thermal contact with each other and heat flows between them until the gases reach a common final temperature (ignore the heat capacity of boxes). Then, the final temperature of the gases, $T_f$, in terms of $T_0$ is

Options:

(A) $T_f = \frac{3}{7} T_0$

(B) $T_f = \frac{7}{3} T_0$

(C) $T_f = \frac{3}{2} T_0$

(D) $T_f = \frac{5}{2} T_0$

Sigma Percentile · Accepted Answer

This problem is a beautiful application of the First Law of Thermodynamics and the concept of thermal equilibrium. It tests your understanding of internal energy, degrees of freedom, and how heat flows between isolated systems.

### Analyzing the Setup

Imagine two rigid boxes placed on a table. Box A contains $1$ mole of Nitrogen ($N_2$), which is a diatomic gas, at an initial temperature of $T_0$. Box B contains $1$ mole of Helium ($He$), a monatomic gas, at a higher initial temperature of $\frac{7}{3}T_0$.

When these two boxes are brought into thermal contact, heat will naturally flow from the hotter body (Box B) to the cooler body (Box A) until they reach a common final equilibrium temperature, let's call it $T_f$.

### The Master Equation

The problem states that the boxes are rigid. This is a crucial constraint! A rigid box means its volume cannot change, so the change in volume $\Delta V = 0$. Consequently, the work done by the gases is zero ($W = P\Delta V = 0$).

Furthermore, the system of the two boxes is isolated from the surroundings. According to the First Law of Thermodynamics, the total internal energy of an isolated system remains constant. Therefore, the sum of the changes in internal energy of the two gases must be zero:

$$\Delta U_A + \Delta U_B = 0$$

### Degrees of Freedom and Heat Capacity

To calculate the change in internal energy, we use the formula $\Delta U = n C_V \Delta T$, where $n$ is the number of moles, $C_V$ is the molar heat capacity at constant volume, and $\Delta T$ is the change in temperature.

We need to determine $C_V$ for both gases. The molar heat capacity depends on the degrees of freedom ($f$) of the gas molecules: $C_V = \frac{f}{2}R$.

- **Nitrogen ($N_2$)** is a diatomic gas. At normal temperatures, it has $3$ translational and $2$ rotational degrees of freedom, giving $f = 5$. Thus, $C_{V,A} = \frac{5}{2}R$.
- **Helium ($He$)** is a monatomic gas. It only has $3$ translational degrees of freedom, giving $f = 3$. Thus, $C_{V,B} = \frac{3}{2}R$.

### Final Calculation

Now, let's substitute everything into our master equation. Both gases have $n = 1$ mole.

$$1 \cdot \left(\frac{5}{2}R\right) (T_f - T_0) + 1 \cdot \left(\frac{3}{2}R\right) \left(T_f - \frac{7}{3}T_0\right) = 0$$

Notice that the term $\frac{R}{2}$ is common to both parts of the equation. We can divide the entire equation by $\frac{R}{2}$ to simplify it beautifully:

$$5(T_f - T_0) + 3\left(T_f - \frac{7}{3}T_0\right) = 0$$

Let's expand the brackets. Watch how the $3$ perfectly cancels out the denominator in the second term:

$$5T_f - 5T_0 + 3T_f - 7T_0 = 0$$

Combining the like terms, we get:

$$8T_f - 12T_0 = 0$$

$$8T_f = 12T_0$$

Finally, solving for the equilibrium temperature $T_f$:

**$$T_f = \frac{12}{8}T_0 = \frac{3}{2}T_0$$**

This elegant result shows how the different heat capacities of monatomic and diatomic gases influence the final equilibrium state. The diatomic gas, having more degrees of freedom, acts as a larger 'energy sink' compared to the monatomic gas.

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Visualized Solution

$Initial State$

$Thermal Equilibrium$

$Internal Energy Change$

$Substituting Values$

$Molar Heat Capacities$

$Equation Setup$

$Simplification$

$Expanding Brackets$

$Final Temperature$

$What if boxes had heat capacity?$

The Sigma Insight: Degree of Freedom and Law of Equipartition of Energy

Analyzing the Setup

The Master Equation

Degrees of Freedom and Heat Capacity

Final Calculation

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Visualized Solution

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

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

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

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

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

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

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

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

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

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } What if boxes had heat capacity?