Reduced Pressure of Real Gas using Clausius Equation given Reduced and Actual Parameters Calculator

Reduced Pressure Formula:

\[ P_{red} = \frac{\left(\frac{[R] \times T_{rg}}{V'_m - b'}\right) - \left(\frac{a}{T_{rg} \times (V'_m + c)^2}\right)}{p} \]

Temperature of Real Gas:

Molar Volume of Real Gas:

m³

Clausius Parameter b:

Clausius Parameter a:

Clausius Parameter c:

Pressure:

Reduced Pressure:

Unit Converter ▲

Unit Converter ▼

From:	To:

1. What is Reduced Pressure of Real Gas?

Definition: Reduced pressure is the ratio of the actual pressure of the fluid to its critical pressure. It is a dimensionless quantity used in thermodynamic calculations.

Purpose: This calculator determines the reduced pressure of a real gas using the Clausius equation of state, which accounts for deviations from ideal gas behavior.

2. How Does the Calculator Work?

The calculator uses the Clausius equation:

\[ P_{red} = \frac{\left(\frac{[R] \times T_{rg}}{V'_m - b'}\right) - \left(\frac{a}{T_{rg} \times (V'_m + c)^2}\right)}{p} \]

Where:

\( P_{red} \) — Reduced pressure (dimensionless)
\( [R] \) — Universal gas constant (8.314 J/mol·K)
\( T_{rg} \) — Temperature of real gas (K)
\( V'_m \) — Molar volume of real gas (m³)
\( b' \) — Clausius parameter b (m³)
\( a \) — Clausius parameter a (Pa·m⁶/K·mol²)
\( c \) — Clausius parameter c (m³)
\( p \) — Pressure (Pa)

Explanation: The equation accounts for molecular size (b'), intermolecular forces (a), and additional volume effects (c) in real gases.

3. Importance of Reduced Pressure Calculation

Details: Reduced properties are crucial for applying the principle of corresponding states, which allows prediction of gas behavior based on critical properties.

4. Using the Calculator

Tips: Enter all required parameters in appropriate units. Temperature must be in Kelvin, volumes in cubic meters, and pressure in Pascals.

5. Frequently Asked Questions (FAQ)

Q1: What are typical values for Clausius parameters?
A: Parameters vary by gas. For example, for CO₂: a ≈ 0.3653 Pa·m⁶/K·mol², b ≈ 4.28×10⁻⁵ m³/mol, c ≈ 4.2×10⁻⁵ m³/mol.

Q2: Why use reduced pressure instead of actual pressure?
A: Reduced properties allow comparison of different gases at similar states relative to their critical points.

Q3: What's the range of validity for this equation?
A: The Clausius equation works best for moderate pressures, typically below 10 times the critical pressure.

Q4: How do I find Clausius parameters for a specific gas?
A: These are typically found in thermodynamic tables or determined experimentally.

Q5: What if I get a negative reduced pressure?
A: Check your inputs. Negative values may indicate either incorrect parameters or conditions where the Clausius model breaks down.