The thermally insulated vessel contains an ideal gas of molecular mass M and ratio of specific heat \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 :

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

$\frac{(γ - 1)}{2 γ R} M v^{2} K$

$\frac{γ M v^{2}}{2 R} K$

$\frac{(γ - 1)}{2 R} M v^{2} K$

$\frac{(γ - 1)}{2 (γ + 1) R} M v^{2} K$

$\frac{1}{2}M{\nu ^2} = {C_v}.\Delta T$

$\frac{1}{2}M{\nu ^2} = \frac{R}{{\gamma - 1}}.\Delta T$

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

This problem involves a thermally insulated vessel containing an ideal gas. The vessel is moving with speed $v$ and is suddenly brought to rest. Since the vessel is insulated, no heat is exchanged with the surroundings. We need to find the increase in temperature of the gas due to this sudden stop.

Key Concept: When the vessel is brought to rest, its kinetic energy is converted into internal energy of the gas. Because the process is adiabatic (no heat loss) and the vessel is rigid (assuming constant volume), this energy transfer results in a temperature increase.

Step 1: Identify the energy conversion.
The initial kinetic energy of the vessel (and the gas inside, since they move together) is given by: $K E = \frac{1}{2} m v^{2}$ where $m$ is the total mass of the gas. This kinetic energy is entirely converted into internal energy of the gas, $Δ U$ .

Step 2: Relate the change in internal energy to temperature change.
For an ideal gas, the change in internal energy at constant volume is given by: $Δ U = n C_{v} Δ T$ where $n$ is the number of moles, $C_{v}$ is the molar specific heat at constant volume, and $Δ T$ is the change in temperature.

Step 3: Equate the energy change.
Since the kinetic energy is converted to internal energy: $\frac{1}{2} m v^{2} = n C_{v} Δ T$

Step 4: Express in terms of given parameters.
The total mass $m$ can be written as $m = n M$ , where $M$ is the molecular mass. Substituting: $\frac{1}{2} n M v^{2} = n C_{v} Δ T$ The $n$ cancels out: $\frac{1}{2} M v^{2} = C_{v} Δ T$ So, $Δ T = \frac{M v^{2}}{2 C_{v}}$

Step 5: Relate $C_{v}$ to $γ$ and $R$ .
For an ideal gas, the molar specific heat at constant volume is related to the gas constant $R$ and the ratio of specific heats $γ$ by: $C_{v} = \frac{R}{γ - 1}$ Substituting this into the equation for $Δ T$ : $Δ T = \frac{M v^{2}}{2 \cdot \frac{R}{γ - 1}} = \frac{M v^{2} (γ - 1)}{2 R}$

Final Answer: The temperature increases by $\frac{(γ - 1)}{2 R} M v^{2} K$ .

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