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

Provided Data:

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

Step 1: Adiabatic Process Equation

For adiabatic compression: $T_2 V_2^{\gamma - 1} = T_1 V_1^{\gamma - 1}$

Step 2: Substitute Values

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

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

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

Step 3: Calculate Final Temperature

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

Step 4: Calculate Temperature Change

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

Final Result:

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