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In engineering thermodynamics, understanding how a system changes from one state to another is crucial. These changes are called thermodynamic processes. Each process describes how properties like pressure (P), volume (V), and temperature (T) vary during the transformation.
Among the fundamental processes, three are especially important for entrance exams and practical applications:
These processes help us analyze engines, refrigerators, and many other devices by simplifying complex state changes into manageable models.
In this section, we will explore each process from first principles, understand their characteristics, visualize their behavior using graphs, and solve typical problems to build strong conceptual clarity.
An isochoric process is one in which the volume of the system remains constant throughout the transformation. The term "isochoric" comes from Greek words meaning "equal volume".
Since volume does not change, the system cannot do any mechanical work on its surroundings, because work in thermodynamics is defined as the force applied over a distance, and volume change represents that distance.
However, pressure and temperature can change. For an ideal gas, pressure is directly proportional to temperature at constant volume, as described by the ideal gas law.
Key points:
Real-world example: Heating a sealed, rigid container of gas (like a pressure cooker) increases pressure as temperature rises, but volume remains fixed.
An isobaric process occurs at constant pressure. The word "isobaric" means "equal pressure".
In this process, the volume of the system changes as temperature changes, following the direct proportionality between volume and temperature at constant pressure.
Since volume changes, the system can do work on its surroundings (or have work done on it), calculated by the pressure multiplied by the volume change.
Key points:
Real-world example: Boiling water in an open pan at atmospheric pressure causes the volume of steam to expand as temperature rises.
An isothermal process is one in which the temperature remains constant throughout the process. "Isothermal" means "equal temperature".
Since temperature is constant, the internal energy of an ideal gas does not change (internal energy depends only on temperature for ideal gases). However, pressure and volume change inversely to maintain the constant temperature, following Boyle's law.
Key points:
\[ W = nRT \ln \frac{V_2}{V_1} \]
Real-world example: Slow compression or expansion of gas in a piston allowing heat exchange with surroundings to keep temperature constant.
| Process | Constant Variable | Variable(s) Changed | Work Done \( W \) | Heat Transfer \( Q \) | Key Equation |
|---|---|---|---|---|---|
| Isochoric | Volume \( V \) | Pressure \( P \), Temperature \( T \) | Zero ( \( W=0 \) ) | Changes internal energy only | \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \) |
| Isobaric | Pressure \( P \) | Volume \( V \), Temperature \( T \) | \( W = P (V_2 - V_1) \) | Heat changes internal energy and does work | \( \frac{V_1}{T_1} = \frac{V_2}{T_2} \) |
| Isothermal | Temperature \( T \) | Pressure \( P \), Volume \( V \) | \( W = nRT \ln \frac{V_2}{V_1} \) | Heat equals work done (no internal energy change) | \( P V = \text{constant} \) |
Step 1: Identify the process type. Since the vessel is rigid, volume is constant -> isochoric process.
Step 2: Use the isochoric relation:
\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \]
Step 3: Rearrange to find \( P_2 \):
\[ P_2 = P_1 \times \frac{T_2}{T_1} \]
Step 4: Substitute values:
\[ P_2 = 200\,000 \times \frac{450}{300} = 200\,000 \times 1.5 = 300\,000\, \text{Pa} = 300\, \text{kPa} \]
Answer: The final pressure is 300 kPa.
Step 1: Recognize the process is isobaric (constant pressure).
Step 2: Use the work formula for isobaric process:
\[ W = P (V_2 - V_1) \]
Step 3: Substitute values (convert kPa to Pa):
\[ P = 100\,000\, \text{Pa}, \quad V_1 = 0.5\, \text{m}^3, \quad V_2 = 1.0\, \text{m}^3 \]
Step 4: Calculate work done:
\[ W = 100\,000 \times (1.0 - 0.5) = 100\,000 \times 0.5 = 50\,000\, \text{J} \]
Answer: Work done by the gas is 50,000 J (or 50 kJ).
Step 1: Identify the process as isothermal (constant temperature).
Step 2: Use the work formula for isothermal process:
\[ W = nRT \ln \frac{V_2}{V_1} \]
Step 3: Substitute known values:
\[ n = 2\, \text{mol}, \quad R = 8.314\, \text{J/mol·K}, \quad T = 300\, \text{K}, \quad V_1 = 0.4\, \text{m}^3, \quad V_2 = 0.2\, \text{m}^3 \]
Step 4: Calculate the natural logarithm:
\[ \ln \frac{0.2}{0.4} = \ln 0.5 = -0.693 \]
Step 5: Calculate work done:
\[ W = 2 \times 8.314 \times 300 \times (-0.693) = -3456\, \text{J} \]
Step 6: Interpretation: Negative work means work is done on the gas (compression).
Step 7: For isothermal process, change in internal energy \( \Delta U = 0 \), so heat transfer \( Q = W \).
Answer: Work done on the gas is 3456 J, and heat removed from the gas is also 3456 J.
Step 1: Initial state:
\( P_1 = 1\, \text{atm} = 101325\, \text{Pa} \), \( V_1 = 0.025\, \text{m}^3 \), \( T_1 = 300\, \text{K} \)
Step 2: Isochoric heating to 600 K (volume constant):
Use \( \frac{P_1}{T_1} = \frac{P_2}{T_2} \)
\[ P_2 = P_1 \times \frac{T_2}{T_1} = 101325 \times \frac{600}{300} = 202650\, \text{Pa} \]
Volume remains \( V_2 = 0.025\, \text{m}^3 \)
Step 3: Isobaric expansion at \( P_2 = 202650\, \text{Pa} \) until volume doubles:
\( V_3 = 2 \times V_2 = 0.05\, \text{m}^3 \)
Use \( \frac{V_2}{T_2} = \frac{V_3}{T_3} \) to find \( T_3 \):
\[ T_3 = T_2 \times \frac{V_3}{V_2} = 600 \times \frac{0.05}{0.025} = 1200\, \text{K} \]
Step 4: Isothermal compression back to initial volume \( V_4 = V_1 = 0.025\, \text{m}^3 \) at \( T_3 = 1200\, \text{K} \):
Use ideal gas law to find \( P_4 \):
\[ P_4 V_4 = P_3 V_3 \implies P_4 = P_3 \times \frac{V_3}{V_4} = 202650 \times \frac{0.05}{0.025} = 405300\, \text{Pa} \]
Step 5: Calculate work done in each process:
\[ W = P_2 (V_3 - V_2) = 202650 \times (0.05 - 0.025) = 202650 \times 0.025 = 5066\, \text{J} \]
\[ W = nRT \ln \frac{V_4}{V_3} = 1 \times 8.314 \times 1200 \times \ln \frac{0.025}{0.05} = 9976 \times (-0.693) = -6915\, \text{J} \]
Step 6: Total work done by the gas:
\[ W_\text{total} = 0 + 5066 - 6915 = -1849\, \text{J} \]
Negative total work means net work done on the gas.
Answer:
Step 1: Since volume is constant, use first law of thermodynamics:
\[ \Delta U = Q - W \]
Work done \( W = 0 \) (isochoric process), so \( Q = \Delta U \).
Step 2: Calculate change in internal energy:
\[ \Delta U = n C_V \Delta T = 4 \times 20.8 \times (600 - 300) = 4 \times 20.8 \times 300 = 24960\, \text{J} \]
Answer: Heat added to the gas is 24,960 J (24.96 kJ).
When to use: When solving problems involving isochoric processes to avoid unnecessary calculations.
When to use: For quick calculation of unknown pressure, volume, or temperature in any process.
When to use: To simplify energy balance calculations and focus on work and heat transfer.
When to use: To avoid unit mismatch errors during work calculation.
When to use: When analyzing process graphs in exam questions.
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