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 asks for the change in temperature of a gas that is adiabatically compressed from an initial volume and temperature to a final volume. The container has thermally non-conducting walls, which confirms the process is adiabatic.

Concept Used:

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

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

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

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

Step-by-Step Solution:

Step 1: List the given initial and final state variables and convert the temperature to Kelvin.

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 thermodynamics, we must use the absolute temperature scale (Kelvin). The conversion is \( T(\text{K}) = T(^\circ\text{C}) + 273 \).

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

Step 2: Set up the adiabatic relation to find the final temperature \( T_2 \).

Using 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 equation and calculate \( T_2 \).

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

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

Next, calculate the ratio of the volumes:

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

Now, substitute these 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 question asks for the change in temperature, which is \( \Delta T = T_2 - T_1 \).

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

The change in temperature is 300 K. This corresponds to option (4).

Answer 2

\( V_1 = 800 cm^3 \) \( V_2 = 200 cm^3 \) \( T_1 = 300 \) 

K For adiabatic: \( TV^{\gamma - 1} = \) constant. 

\( T_1 V_1^{\gamma - 1} = T_2 V_2^{\gamma - 1} \) \( (300) (800)^{1.5 - 1} = T_2 (200)^{1.5 - 1} \) \( T_2 = 300 \left( \frac{800}{200} \right)^{0.5} \) \( T_2 = 300 (4)^{1/2} \) 

\( T_2 = 600 \) K \( \Delta T = 600 - 300 = 300 \) K