Question

An ideal gas initially at 0°C temperature, is compressed suddenly to one fourth of its volume. If the ratio of specific heat at constant pressure to that at constant volume is \( \frac{3}{2} \), the change in temperature due to the thermodynamics process is         K.

Hint:

The change in temperature can be derived using the thermodynamic relation between pressure, volume, and temperature for adiabatic processes. Use the initial and final conditions to calculate the change.

Answer 1

Given Data: 

Initial temperature, $T_1 = 0^\circ C = 273 \text{ K}$ 
Final volume, $V_2 = \dfrac{V_1}{4}$ 
Ratio of specific heats, $\gamma = \dfrac{C_p}{C_v} = \dfrac{3}{2}$

Step 1: Relation for adiabatic process

For a sudden (adiabatic) compression, $T_2 V_2^{\gamma - 1} = T_1 V_1^{\gamma - 1}$

Step 2: Substitute given values

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

$\displaystyle \frac{T_2}{273} = \left(\frac{V_1}{V_1/4}\right)^{\frac{3}{2} - 1}$ 

$\displaystyle \frac{T_2}{273} = 4^{\frac{1}{2}} = 2$

Step 3: Calculate final temperature

$T_2 = 2 \times 273 = 546 \text{ K}$

Step 4: Change in temperature

$\Delta T = T_2 - T_1 = 546 - 273 = 273 \text{ K}$

Final Answer:

$\boxed{273\ \text{K}}$

Answer 2

To determine the temperature change during an adiabatic compression of an ideal gas, we follow these steps:

1. Initial Conditions:
- Initial temperature (T₁) = 0°C = 273 K
- Initial volume (V₁)
- Final volume (V₂) = V₁/4
- Ratio of specific heats (γ) = Cₚ/Cᵥ = 3/2

2. Adiabatic Process Relation:
For an adiabatic process, we use:

\[ T V^{\gamma - 1} = \text{constant} \]
Which gives the relation:

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

3. Calculate Final Temperature:
Solving for the final temperature T₂:

\[ T_2 = T_1 \left( \frac{V_1}{V_2} \right)^{\gamma - 1} = 273 \left( \frac{V_1}{V_1/4} \right)^{1/2} = 273 \times (4)^{1/2} = 273 \times 2 = 546 \text{ K} \]

4. Determine Temperature Change:
The change in temperature is:

\[ \Delta T = T_2 - T_1 = 546 \text{ K} - 273 \text{ K} = 273 \text{ K} \]

Key Points:
- The process is adiabatic (no heat transfer)
- Volume reduces to 1/4th of original
- Temperature doubles due to the compression
- γ = 3/2 indicates a monatomic ideal gas

Final Answer:
The temperature increases by \(\boxed{273 \text{ K}}\).