Question 12

A thermally insulated vessel contains an ideal gas of molecular mass $$M$$ and ratio of specific heats $$\gamma$$. It is moving with speed $$v$$ and is suddenly brought to rest. Assuming no heat is lost to the surroundings, its temperature increases by:

Solution

Solution & Explanation

1. Relate Bulk Kinetic Energy to Internal Thermal Energy

When the moving vessel is suddenly brought to rest, its entire macroscopic bulk kinetic energy ($$K_{\text{bulk}}$$) is stopped. Since the vessel is completely thermally insulated, no heat escapes to the environment ($$\Delta Q = 0$$). According to the First Law of Thermodynamics, this lost mechanical energy is converted entirely into the internal microscopic thermal energy ($$\Delta U$$) of the gas molecules:

$$\Delta U = K_{\text{bulk}}$$

Let $$m$$ be the total mass of the gas contained within the vessel. The bulk kinetic energy lost during the sudden braking phase is given by:

$$K_{\text{bulk}} = \frac{1}{2} \cdot m \cdot v^2$$

2. Express Internal Energy Change in terms of Specific Heat

The increase in internal energy ($$\Delta U$$) for $$n$$ moles of an ideal gas experiencing a temperature rise of $$\Delta T$$ is defined as:

$$\Delta U = n \cdot C_v \cdot \Delta T$$

Where:

$$n$$ = Number of moles of the gas, which can be rewritten as total mass divided by molecular mass ($$n = \frac{m}{M}$$).
$$C_v$$ = Molar specific heat capacity at constant volume.

Using the standard thermodynamic relation linking $$C_v$$ to the specific heat ratio ($$\gamma = \frac{C_p}{C_v}$$) and the universal gas constant ($$R$$):

$$C_v = \frac{R}{\gamma - 1}$$

Substituting both $$n$$ and $$C_v$$ expressions back into our internal energy formula yields:

$$\Delta U = \left(\frac{m}{M}\right) \cdot \left(\frac{R}{\gamma - 1}\right) \cdot \Delta T$$

3. Equate the Energy Terms and Isolate Temperature Rise ($\Delta T$)

Equating the macroscopic kinetic energy loss to the microscopic internal energy gain:

$$\frac{1}{2} \cdot m \cdot v^2 = \frac{m \cdot R \cdot \Delta T}{M \cdot (\gamma - 1)}$$

We can cancel out the total gas mass variable ($$m$$) from both sides of the balance equation:

$$\frac{1}{2} \cdot v^2 = \frac{R \cdot \Delta T}{M \cdot (\gamma - 1)}$$

Isolating the temperature increment variable ($$\Delta T$$) yields:

$$\Delta T = \frac{(\gamma - 1) \cdot M \cdot v^2}{2R} \,\, \text{K}$$

Concept Check: Because the vessel's organized kinetic energy is randomized into chaotic molecular motions upon stopping, the temperature rise scales directly with the molecular weight ($$M$$). Heavier molecules pack more mechanical momentum per mole for a given speed, resulting in a higher internal energy spike when stopped.