Question

Superheated steam at 1500 kPa, has a specific volume of 2.75 m\(^3\)/kmol and compressibility factor (Z) of 0.95. The temperature of steam is________°C (round off to the nearest integer).

• A. 522
• B. 471
• C. 249
• D. 198

Hint:

When solving for the temperature of superheated steam, remember to adjust the ideal gas equation by incorporating the compressibility factor \( Z \) and use appropriate unit conversions for gas constants.

Answer 1

The Correct Option is C

We are given the following data:
- Pressure \( P = 1500 \, \text{kPa} \),
- Specific volume \( v = 2.75 \, \text{m}^3/\text{kmol} \),
- Compressibility factor \( Z = 0.95 \).
We will use the ideal gas equation, modified with the compressibility factor, to solve for the temperature: \[ PV = ZnRT. \] Rearranging for temperature \( T \), we get: \[ T = \frac{P \cdot v}{Z \cdot R}, \] where:
- \( R = 8.314 \, \text{J/mol·K} \) (the universal gas constant),
- \( P \) and \( v \) are in the appropriate units.
Since \( P = 1500 \, \text{kPa} = 1500 \times 10^3 \, \text{Pa} \) and \( v = 2.75 \, \text{m}^3/\text{kmol} \), we use the ideal gas constant in kPa·m\(^3\)/kmol·K, which is \( R = 8.314 \times 10^{-3} \, \text{kPa·m}^3/\text{kmol·K} \). Now, substituting the values: \[ T = \frac{1500 \times 2.75}{0.95 \times 8.314 \times 10^{-3}}. \] After calculation: \[ T = 249.3 \, \text{°C}. \] Thus, the temperature is 249°C, which corresponds to Option (C).
Final Answer: (C) 249