Question

A thermally insulated vessel contains an ideal gas of molecular mass M and ratio of specific heats 1.4. Vessel is moving with speed ν and is suddenly brought to rest. Assuming no heat is lost to the surrounding and vessel temperature of the gas increases by(R = universal gas constant

• A. \(\frac{Mv^2}{7R}\)
• B. \( \frac{Mv^2}{5R} \)
• C. \(\frac{2Mv^2}{7R}\)
• D. \(\frac{7Mv^2}{5R}\)

Answer 1

The Correct Option is B

 To solve this problem, let's analyze the situation step by step. We are given an ideal gas in a thermally insulated vessel, meaning no heat is exchanged with the environment. The vessel was moving with a speed \(\nu\) and is suddenly brought to rest. As a result, the kinetic energy of the vessel gets converted into the internal energy of the gas, leading to a rise in its temperature. Let's find the increase in temperature of the gas.

Given:

• Ratio of specific heats, \(\gamma = 1.4\).
• Molecular mass of the gas, \(M\).
• Vessel speed, \(\nu\).
• Universal gas constant, \(R\).

 

Firstly, the kinetic energy of the vessel before it stops is given by: \(\frac{1}{2}Mv^2\)

This kinetic energy is converted entirely into the internal energy of the gas due to the insulation of the vessel.

The change in internal energy, \(\Delta U\), for an ideal gas due to a temperature change \(\Delta T\) is given by: \(\Delta U = nC_V\Delta T\)

Where:

• \(n\) is the number of moles,
• \(C_V\) is the molar specific heat at constant volume.

 

For a monoatomic gas, using \(C_V\), we have the relation: \(C_V = \frac{R}{\gamma - 1}\)

\(C_V\) for our gas will be: \(\frac{R}{1.4 - 1} = \frac{R}{0.4} = \frac{5R}{2}\)

Now, equating the kinetic energy to the change in internal energy: \(\frac{1}{2}Mv^2 = nC_V\Delta T\)

Since number of moles \(n\) can be represented as mass \(m\) divided by molar mass \(M\), i.e., \(n = \frac{m}{M}\), the equation becomes: \(\frac{1}{2}Mv^2 = \frac{m}{M} \cdot \frac{5R}{2} \cdot \Delta T\)

Thus, solving for \(\Delta T\): \(\Delta T = \frac{Mv^2}{5R}\)

So the temperature increase of the gas is given by \(\frac{Mv^2}{5R}\).

Therefore, the correct answer is: \(\frac{Mv^2}{5R}\), which matches option \(B\).