Question

A gas is kept in a container having walls which are thermally non-conducting. Initially the gas has a volume of 800 $ cm^3 $ and temperature 27°C. The change in temperature when the gas is adiabatically compressed to 200 $ cm^3 $ is: (Take $ \gamma $ = 1.5 : $ \gamma $ is the ratio of specific heats at constant pressure and at constant volume)

• A. 327 K
• B. 600 K
• C. 522 K
• D. 300 K

Hint:

Use the adiabatic process equation to relate the initial and final temperatures and volumes. Remember to use consistent units for volume.

Answer 1

The Correct Option is D

The problem requires determining the temperature variation of a gas subjected to an adiabatic compression, transitioning from an initial volume and temperature to a final volume. The presence of thermally non-conducting walls in the container confirms the adiabatic nature of the process.

Concept Used:

For a reversible adiabatic process, the relationship governing the absolute temperature (T) and volume (V) of a gas is described by the equation:

\[ TV^{\gamma-1} = \text{constant} \]

Here, \( \gamma \) represents the ratio of specific heats (\( C_p/C_v \)). For two distinct states, initial (1) and final (2), this relationship can be expressed as:

\[ T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1} \]

Step-by-Step Solution:

Step 1: Enumerate the provided initial and final state variables and convert the temperature to Kelvin.

The given values are:

• Initial Volume, \( V_1 = 800 \, \text{cm}^3 \)
• Final Volume, \( V_2 = 200 \, \text{cm}^3 \)
• Ratio of specific heats, \( \gamma = 1.5 \)
• Initial Temperature, \( T_1 = 27^\circ \text{C} \)

In thermodynamic calculations, it is imperative to utilize the absolute temperature scale (Kelvin). The conversion formula is \( T(\text{K}) = T(^\circ\text{C}) + 273 \).

\[T_1 = 27 + 273 = 300 \, \text{K}\]

Step 2: Formulate the adiabatic relation to derive the final temperature \( T_2 \).

By rearranging the relation \( T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1} \), we can solve for \( T_2 \):

\[T_2 = T_1 \left( \frac{V_1}{V_2} \right)^{\gamma-1}\]

Step 3: Substitute the given values into the derived equation and compute \( T_2 \).

Initially, calculate the exponent \( \gamma - 1 \):

\[\gamma - 1 = 1.5 - 1 = 0.5 = \frac{1}{2}\]

Subsequently, compute the ratio of the volumes:

\[\frac{V_1}{V_2} = \frac{800 \, \text{cm}^3}{200 \, \text{cm}^3} = 4\]

Now, substitute these computed values into the equation for \( T_2 \):

\[T_2 = 300 \, \text{K} \times (4)^{0.5} = 300 \, \text{K} \times \sqrt{4}\]\[T_2 = 300 \, \text{K} \times 2 = 600 \, \text{K}\]

Final Computation & Result:

The problem specifically asks for the change in temperature, calculated as \( \Delta T = T_2 - T_1 \).

\[\Delta T = 600 \, \text{K} - 300 \, \text{K} = 300 \, \text{K}\]

The resulting change in temperature is 300 K. This aligns with option (4).