Chapter 6.6, Problem 26E

To determine

To Calculate: The work done by steam while expanding between two given volumes by a process that follows a certain $P-V$ relation.

Answer to Problem 26E

The work done by the gas in expanding from $100i{n}^3$ to $800i{n}^3$ is $9035.8 lb\cdot in$

Explanation of Solution

Given Information:

• The $P-V$ relation is given by $P{V}^{1.4}=k$ .
• The initial pressure and volume are $160lb/i{n}^2$ and $100i{n}^3$ respectively.
• The final volume is $800i{n}^3$

Concept Used: If $P\left(V\right)$ is the pressure of a gas at volume $V$ , then the work done by the gas in expanding from volume $V_1$ to $V_2$ is given by $W_{V_1\to V_2}={{\displaystyle \int }}_{V_1}^{V_2}P\left(V\right)dV$

Put the values of initial pressure and volume in the $P-V$ relation to find the value of $k$ as

  $k=P_1\cdot V_1^{1.4}=160\times {100}^{1.4}=100953$

From the calculated value of $k$ , the expression for pressure $P$ of the gas as a function its volume $V$ is

  $P\left(V\right)=\frac{k}{V^{1.4}}=\frac{100953}{V^{1.4}}$

Put the expression for $P\left(V\right)$ in the work equation of a gas to get

  $W_{100\to 800}={{\displaystyle \int }}_{100}^{800}\frac{100953}{V^{1.4}}dV$

  $={\frac{-100953}{V^{0.4}}]}_{100}^{800}=-100953\left(\frac{1}{{800}^{0.4}}-\frac{1}{{100}^{0.4}}\right)$

  $=9035.78$

So, the work done by the gas in expanding from $100i{n}^3$ to $800i{n}^3$ is $9035.8 lb\cdot in$