Question

During some actual expansion and compression processes in piston-cylinder devices, the gases have been observed to satisfy the relationship \(P V^{n}=C\) where \(n\) and \(C\) are constants. Calculate the work done when a gas expands from \(350 \mathrm{kPa}\) and \(0.03 \mathrm{m}^{3}\) to a final volume of \(0.2 \mathrm{m}^{3}\) for the case of \(n=1.5\).

Short answer

Answer: The work done during the gas expansion is approximately \(-43257.48\,\text{J}\).

Step-by-step solution

Understanding the given relationship and parameters

We are given the relationship between pressure and volume: \(PV^n = C\), where \(P\) is pressure, \(V\) is volume, and \(n\) and \(C\) are constants. We are also given the initial pressure \(P_1 = 350\,\text{kPa}\), initial volume \(V_1 = 0.03\,\text{m}^3\), final volume \(V_2 = 0.2\,\text{m}^3\), and \(n = 1.5\).

Determine the constant C

Using the initial pressure and volume, we can calculate the constant \(C\). From the relationship \(P_1 V_1^{n} = C\), we plug in the given values: \(C = (350\,\text{kPa})(0.03\,\text{m}^3)^{1.5}\) Converting kPa to Pa, we have: \(C = (350000\,\text{Pa})(0.03\,\text{m}^3)^{1.5}\) Now, calculate the value of \(C\). \(C \approx 113329.48\,\text{Pa}\cdot\text{m}^{3n}\)

Determine the final pressure P2

With the value of \(C\), we can determine the final pressure \(P_2\). We have \(P_2 V_2^{n} = C\). Plug in the values: \(P_2 \times (0.2\,\text{m}^3)^{1.5} = 113329.48\,\text{Pa}\cdot\text{m}^{3n}\) Now, solve for \(P_2\). \(P_2 \approx 40872.94\,\text{Pa}\)

Calculate the work done during expansion

The work done during the expansion of the gas in a piston-cylinder device can be derived by integrating the relationship \(PV^n = C\) with respect to volume. The work done during the process is given by: \(W = \int_{V_1}^{V_2} P\, dV = \int_{V_1}^{V_2} \frac{C}{V^n}\, dV\) Plug in the values of \(n\) and \(C\): \(W = \int_{0.03\,\text{m}^3}^{0.2\,\text{m}^3} \frac{113329.48\,\text{Pa}\cdot\text{m}^{3n}}{V^{1.5}}\, dV\) Now, integrate the expression with respect to volume: \(W \approx \left[-\frac{2}{3} \times 113329.48\,\text{Pa}\cdot\text{m}^{3n} V^{-0.5}\right]_{0.03\,\text{m}^3}^{0.2\,\text{m}^3}\) When we apply the limits and calculate the result, we get: \(W \approx -43257.48\,\text{J}\)

Interpret the result

The result for the work done during the expansion is negative, which indicates that the gas has performed work on its surroundings while expanding. So, the work done during the gas expansion from \(350\,\text{kPa}\) and \(0.03\,\text{m}^{3}\) to a final volume of \(0.2\,\text{m}^{3}\) for the case of \(n=1.5\) is approximately \(-43257.48\,\text{J}\).