Question

At an initial pressure of 100 kPa, 1 kg air is compressed reversibly from 15 litres to 1 litre at final pressure of 2000 kPa. Neglecting other losses, the Polytropic Exponent (n) is ________. (Rounded off to three decimal places)

Hint:

Use P1 V1^n equals P2 V2^n and solve for n from the pressure and volume ratios.

Answer 1

Correct Answer: 1.106

Step 1: Write the polytropic process relation.
For a reversible polytropic compression, \(P_1 V_1^{n} = P_2 V_2^{n}\), where \(P_1, V_1\) are the initial pressure and volume and \(P_2, V_2\) are the final values.
Taking natural logs of both sides gives \(n = \dfrac{\ln(P_2/P_1)}{\ln(V_1/V_2)}\).

Step 2: Substitute the given values.
\(P_1 = 100\) kPa, \(P_2 = 2000\) kPa, \(V_1 = 15\) L, \(V_2 = 1\) L.
\[ n = \frac{\ln(2000/100)}{\ln(15/1)} = \frac{\ln 20}{\ln 15} \]
\[ n = \frac{2.9957}{2.7081} \]

Step 3: Compute the numeric value.
\[ n \approx 1.106 \]
The mass of air (1 kg) is not needed here, since the polytropic exponent depends only on the pressure and volume ratios.

Final Answer:
The polytropic exponent is about 1.106, which falls between 1.100 and 1.111 as expected. \[ \boxed{n \approx 1.106} \]