Question

A piston-cylinder assembly having 250 mm diameter contains 0.01 kg of water vapor at 1 MPa and 200 °C. The specific volume of the vapor is 0.20602 m³/kg. The system expands as per the relation \(pV^n = \text{constant}\), where \(n\) is the polytropic exponent. The expansion of water vapour displaces the piston by 50 mm. If the final pressure is 0.35 MPa, the value of the exponent \(n\) is ................ (rounded off to two decimal places).

Hint:

For polytropic processes, always use: \[ n = \frac{\ln(p_1/p_2)}{\ln(V_2/V_1)} \] This formula is the most direct way without intermediate steps.

Answer 1

Correct Answer: 1.3

Step 1: Initial volume.
\[ V_1 = m \cdot v = (0.01)(0.20602) = 0.0020602 \, m^3 \]

Step 2: Area of piston.
\[ A = \frac{\pi d^2}{4} = \frac{\pi (0.25)^2}{4} = 0.0491 \, m^2 \]

Step 3: Displacement volume.
\[ \Delta V = A \cdot \Delta x = 0.0491 \times 0.05 = 0.002455 \, m^3 \] \[ V_2 = V_1 + \Delta V = 0.0020602 + 0.002455 = 0.004515 \, m^3 \]

Step 4: Polytropic relation.
\[ p_1 V_1^n = p_2 V_2^n \] \[ n = \frac{\ln(p_1/p_2)}{\ln(V_2/V_1)} \] \[ n = \frac{\ln(1.0/0.35)}{\ln(0.004515/0.0020602)} \] \[ n = \frac{\ln(2.857)}{\ln(2.19)} = \frac{1.049}{0.799} = 1.31 \] Final Answer:
\[ \boxed{1.31} \]