Chapter 19, Problem 74PQ

A gas is in a container of volume V₀ at pressure P₀. It is being pumped out of the container by a piston pump. Each stroke of the piston removes a volume V_s through valve A and then pushes the air out through valve B as shown in Figure P19.74. Derive an expression that relates the pressure P_n of the remaining gas to the number of strokes n that have been applied to the container.

FIGURE P19.74

To determine

The expression that relates the pressure $P_n$ of the remaining gas to the number of strokes $n$ that has been applied to the container.

Answer to Problem 74PQ

The expression that relates the pressure $P_n$ of the remaining gas to the number of strokes $n$ that has been applied to the container is $P_n=P_0{\left(\frac{V_0}{V_0+V_s}\right)}^n$.

Explanation of Solution

In each stroke, a constant volume of gas is removed from the container of volume $V_0$ . Thus in each stroke, number of molecules decreases as the pressure decreases. The process takes place at constant temperature and Boyle’s law is applicable.

Consider the case when valve A opens and B closes and the gas expands, reaching a new density.

At this stage number of molecules remains constant. After this stage, valve A closes and B opens, and a volume $V_s$ is removed from the gas.

Write Boyles law.

  $PV=\text{constant}$

Here, $P$ is the pressure of the gas and $V$ is the volume of the gas.

Let $P_0$ be the pressure of the gas in container of volume $V_0$, $P_1$ is the pressure of the gas after first stroke and volume expands to $V_0+V_s$ .

Apply Boyle’s law after the first stroke.

  $P_0{V}_0=P_1\left(V_0+V_s\right)$                                                                                                                 (I)

Here, $P_0$ is the pressure of the gas before first stroke, $V_0$ is the volume of the container, $P_1$ is the pressure of the gas after first stroke, $V_s$ is the volume removed from the gas.

Rearrange equation (I) to get $P_1$.

  $P_1=P_0\frac{V_0}{V_0+V_s}$                                                                                                               (II)

When the gas is expelled, volume $V_s$ is removed and there is again a volume $V_0$, now at pressure $P_1$ , which expands in the same way as in the previous stroke.

Apply Boyle’s law for second stroke.

  $P_1{V}_0=P_2\left(V_0+V_s\right)$                                                                                                              (III)

Here, $P_2$ is the pressure of the gas after second stroke.

Rearrange equation (III) to get $P_2$ .

  $P_2=P_1\frac{V_0}{V_0+V_s}$                                                                                                             (IV)

Substitute (II) in (IV) to get relation between $P_2$ and $P_0$ .

  $\begin{array}{c}{P}_2=\left(P_0\frac{V_0}{V_0+V_s}\right)\frac{V_0}{V_0+V_s}\\ =P_0{\left(\frac{V_0}{V_0+V_s}\right)}^2\end{array}$                                                                                                    (V)

Similarly write the expression for pressure of the gas after ${\left(n-1\right)}^{\text{th}}$ stroke.

  $P_{n-1}=P_0{\left(\frac{V_0}{V_0+V_s}\right)}^{n-1}$                                                                                                         (VI)

In general continuing the pattern, the pressure of the remaining gas will be $P_n$ .

Apply Boyle’s law after $n^{\text{th}}$ stroke.

  $P_{n-1}{V}_0=P_n\left(V_0+V_s\right)$                                                                                                         (VII)

Here, $P_{n-1}$ pressure of the gas before $n^{\text{th}}$ stroke and $P_n$ is the pressure of the gas after $n^{\text{th}}$ stroke.

Rearrange above equation to get $P_n$ .

  $P_n=P_{n-1}\left(\frac{V_0}{V_0+V_s}\right)$                                                                                                         (VIII)

Substitute (VI) in (VIII) to get relation between $P_n$ and $P_0$ .

  $\begin{array}{c}{P}_n=\left(P_0{\left(\frac{V_0}{V_0+V_s}\right)}^{n-1}\right)\left(\frac{V_0}{V_0+V_s}\right)\\ =P_0{\left(\frac{V_0}{V_0+V_s}\right)}^n\end{array}$                                                                                      (IX)

Conclusion:

Therefore, the expression that relates the pressure $P_n$ of the remaining gas to the number of strokes $n$ that has been applied to the container is $P_n=P_0{\left(\frac{V_0}{V_0+V_s}\right)}^n$.