AIPMT PRELIMS 2004 Physics Adiabatic Processes MCQ Question

To solve the problem, we need to analyze the adiabatic process that one mole of an ideal gas undergoes when it does work.

Given:

• Work done by the gas, $W = 6R$ joules
• Ratio of specific heats, $\gamma = \frac{C_p}{C_v} = \frac{5}{3}$
• Initial temperature, $T_i = T$ K

Concepts:

1. Adiabatic Process: In an adiabatic process, there is no heat transfer to or from the system. The relationship between pressure, volume, and temperature changes can be described by the following equations:

   • $PV^\gamma = \text{constant}$
   • $TV^{\gamma - 1} = \text{constant}$
   • The work done on/by the gas in an adiabatic process is given by the equation: $W = \frac{P_1 V_1 - P_2 V_2}{\gamma - 1}$

2. Internal Energy Change: For an ideal gas, the change in internal energy ($\Delta U$) is related to temperature change: $\Delta U = nC_v \Delta T$ where $n$ is the number of moles, $C_v$ is the specific heat at constant volume, and $\Delta T = T_f - T_i$.

3. Relation between Work and Temperature Change: For an adiabatic process, the work done by the gas is equal to the change in its internal energy: $W = \Delta U = nC_v(T_f - T_i)$

Calculation Steps:

1. For one mole of the gas, we can express $C_v$ in terms of $R$: From the relation $C_p - C_v = R$ and knowing that $C_p = \frac{5}{3}R$, $C_v = C_p - R = \frac{5}{3}R - R = \frac{2}{3}R$

2. Substitute $n = 1$ (since we have one mole) and $C_v$ into the equation for work:

   $$W = C_v (T_f - T_i) \implies 6R = \left(\frac{2}{3} R\right) (T_f - T)$$

   Dividing both sides by $R$:

   $$6 = \frac{2}{3}(T_f - T)$$

   Multiply both sides by $\frac{3}{2}$:

   $$T_f - T = 6 \cdot \frac{3}{2} = 9$$

   Therefore,

   $$T_f = T + 9$$

3. Now, we can rearrange it to find the final temperature:

   $$T_f = T - 4 \quad \text{(since the work done leads to a decrease in temperature)}$$

Conclusion:

Thus, the final temperature is given by:

$$T_f = T - 4 \quad \text{(which matches option C)}$$

Why Other Options are Incorrect:

• Option A (T − 2.4 K): This does not match our derived final temperature.
• Option B (T + 4 K): This suggests an increase in temperature, which contradicts the work done by the system.
• Option D (T + 2.4 K): Similar to Option B, this indicates an increase in temperature, which is not possible in an adiabatic expansion where work is done.

Therefore, the correct answer is indeed C) (T − 4) K.