A cylinder of 0.9 m³ in volume contains air at 0°C and 39.24 N/cm² absolute pressure. The air is compressed to 0.45 m³. Find (i) the pressure inside the cylinder assuming an isothermal process, and (ii) the pressure and temperature assuming an adiabatic process.

Isothermal and Adiabatic Compression

Problem Statement

A cylinder of 0.9 m³ in volume contains air at 0°C and 39.24 N/cm² absolute pressure. The air is compressed to 0.45 m³. Find (i) the pressure inside the cylinder assuming an isothermal process, and (ii) the pressure and temperature assuming an adiabatic process. Take k = 1.4 for air.

Given Data

Initial Volume, $V_1 = 0.9 \, \text{m}^3$
Initial Temperature, $t_1 = 0^\circ\text{C}$
Initial Pressure, $p_1 = 39.24 \, \text{N/cm}^2$
Final Volume, $V_2 = 0.45 \, \text{m}^3$
Adiabatic Index, $k = 1.4$

Solution

1. Convert Initial Conditions to SI Units

Temperature Conversion to Kelvin:

$$ T_1 = t_1(^\circ\text{C}) + 273.15 $$ $$ T_1 = 0 + 273.15 = 273.15 \, \text{K} $$

Pressure Conversion to N/m² (Pascals):

$$ p_1 = 39.24 \, \frac{\text{N}}{\text{cm}^2} \times \left(\frac{100 \, \text{cm}}{1 \, \text{m}}\right)^2 $$ $$ p_1 = 39.24 \times 10^4 \, \text{N/m}^2 $$

(i) Isothermal Process (Constant Temperature)

For an isothermal process, $p_1 V_1 = p_2 V_2$.

$$ p_2 = p_1 \left(\frac{V_1}{V_2}\right) $$ $$ p_2 = (39.24 \times 10^4 \, \text{N/m}^2) \times \left(\frac{0.9 \, \text{m}^3}{0.45 \, \text{m}^3}\right) $$ $$ p_2 = (39.24 \times 10^4) \times 2 $$ $$ p_2 = 78.48 \times 10^4 \, \text{N/m}^2 = 78.48 \, \text{N/cm}^2 $$

The temperature remains constant: $t_2 = t_1 = 0^\circ\text{C}$.

(ii) Adiabatic Process (No Heat Transfer)

Final Pressure Calculation ($p_1 V_1^k = p_2 V_2^k$):

$$ p_2 = p_1 \left(\frac{V_1}{V_2}\right)^k $$ $$ p_2 = (39.24 \times 10^4 \, \text{N/m}^2) \times \left(\frac{0.9}{0.45}\right)^{1.4} $$ $$ p_2 = (39.24 \times 10^4) \times (2)^{1.4} $$ $$ p_2 \approx (39.24 \times 10^4) \times 2.639 $$ $$ p_2 \approx 103.55 \times 10^4 \, \text{N/m}^2 = 103.55 \, \text{N/cm}^2 $$

Final Temperature Calculation ($T_1 V_1^{k-1} = T_2 V_2^{k-1}$):

$$ T_2 = T_1 \left(\frac{V_1}{V_2}\right)^{k-1} $$ $$ T_2 = 273.15 \, \text{K} \times \left(\frac{0.9}{0.45}\right)^{1.4-1.0} $$ $$ T_2 = 273.15 \times (2)^{0.4} $$ $$ T_2 \approx 273.15 \times 1.3195 $$ $$ T_2 \approx 360.42 \, \text{K} $$

Converting the final temperature back to Celsius:

$$ t_2 = T_2 - 273.15 $$ $$ t_2 = 360.42 - 273.15 $$ $$ t_2 \approx 87.27^\circ\text{C} $$

Final Results:

Isothermal Process:
Final Pressure $ p_2 = 78.48 \, \text{N/cm}^2 $

Adiabatic Process:
Final Pressure $ p_2 \approx 103.55 \, \text{N/cm}^2 $
Final Temperature $ t_2 \approx 87.27^\circ\text{C} $

Explanation of Processes

1. Isothermal Process:
This is a thermodynamic process that occurs at a constant temperature. For the temperature to remain constant during compression, any heat generated by the work done on the gas must be slowly removed from the system. It represents an idealized, very slow compression.

2. Adiabatic Process:
This is a process that occurs with no heat transfer between the system (the air) and its surroundings. This happens if the process is perfectly insulated or occurs very rapidly, leaving no time for heat to escape. The work done on the gas increases its internal energy, which causes a rise in both its temperature and pressure.

Physical Meaning and Comparison

The results clearly show that for the same volume change, the adiabatic compression results in a much higher final pressure and temperature than the isothermal compression.

This is because in the adiabatic case, the energy from the work of compression is trapped inside the gas, increasing its internal energy and temperature. This temperature increase provides an additional "push," causing the pressure to rise more significantly than it would from the volume reduction alone.

In the real world, fast compression, like in an engine cylinder or when using a manual tire pump, is nearly adiabatic, which is why the pump and the compressed air get hot. A very slow, non-insulated compression would more closely resemble an isothermal process.