Question

A gas of mass 'm' and molecular weight 'M' is flowing in an insulated tube with a velocity '2V'. If the flow of the gas is suddenly stopped and all the kinetic energy is utilized to compress the gas, the increase in the temperature of the gas is ($\gamma$ is ratio of specific heats, R is universal gas constant)

• A. $\frac{2MV^2 (\gamma - 1)}{R}$
• B. $\frac{mV^2 (\gamma - 1)}{2MR}$
• C. $\frac{mV^2 \gamma}{2R}$
• D. $\frac{MV^2 \gamma}{2R}$

Hint:

In processes where kinetic energy is converted to internal energy, remember to relate the change in kinetic energy to the change in internal energy using the first law of thermodynamics.

Answer 1

The Correct Option is A

Step 1: Calculate the initial kinetic energy of the gas.
$KE_1 = \frac{1}{2} m (2V)^2 = 2mV^2$
Step 2: Calculate the change in kinetic energy.
The final kinetic energy is $KE_2 = 0$ (flow is stopped). $\Delta KE = KE_2 - KE_1 = 0 - 2mV^2 = -2mV^2$
Step 3: Apply the first law of thermodynamics for an adiabatic process.
Since the tube is insulated ($Q = 0$) and the work done on the gas is equal to the negative of the change in kinetic energy ($W = -\Delta KE = 2mV^2$), we have: $\Delta U = Q + W = 0 + 2mV^2 = 2mV^2$
Step 4: Relate the change in internal energy to the change in temperature.
$\Delta U = n C_V \Delta T$, where $n = \frac{m}{M}$ and $C_V = \frac{R}{\gamma - 1}$. $\Delta U = \frac{m}{M} \frac{R}{\gamma - 1} \Delta T$
Step 5: Equate the two expressions for $\Delta U$.
$2mV^2 = \frac{m}{M} \frac{R}{\gamma - 1} \Delta T$
Step 6: Solve for the increase in temperature $\Delta T$.
$\Delta T = \frac{2mV^2 M (\gamma - 1)}{mR} = \frac{2MV^2 (\gamma - 1)}{R}$
Thus, the increase in the temperature of the gas is $ \boxed{\frac{2MV^2 (\gamma - 1)}{R}} $.