Chapter 13, Problem 85GP

To determine

To Find: The time for which cylinder will last.

Answer to Problem 85GP

The time up to which the cylinder last for a given flow rate is $909\min$ .

Explanation of Solution

Given info:

Gauge pressure, $P_1=2000\text{ psi }=13800\text{ kPa }=1.3\times {10}^7\text{ Pa}$ .

Volume, $V_1=16\text{ L }=0.016{\text{ m}}^{\text{3}}$ .

Temperature, $T_1=295\text{ K}$ .

Flow rate, $Q=2.4\text{ L/min }=\frac{2.4\times {10}^{-3}}{60}{\text{ m}}^{\text{3}}\text{/s}$ .

Cylinder giving out oxygen to some hospital at a rate of 2.4 lit/min at following condition.

Gauge pressure, $P_2=1\text{ atm }=1.013\times {10}^5\text{ Pa}$ .

Volume, $V_2=2.4\text{ L }=2.4\times {10}^{-3}{\text{ m}}^{\text{3}}$ .

Temperature, $T_2=295\text{ K}$ .

Formula used:

Write the expression for ideal gas law.

  $P\times V=n\times R\times T$

Write the expression for number of molecule escape per minute.

  $n=\frac{P\times V}{R\times T}$

Here, $P$ is the pressure of oxygeninside the cylinder,

  $V$ is the volume of oxygen in the cylinder,

  $n$ is the number of molecules of oxygen,

  $R$ is the universal gas constant,

  $T$ is the Temperature of the oxygen in the cylinder.

Calculation:

Substitute the given values in the above equations,

For the Volume of $16L$ , number of molecules are

  $n=\frac{1.3\times {10}^7\times 0.016}{8.314\times 295}=90\text{ moles}$

For the Volume of 2.4 L, number of molecules are

  $n=\frac{1.013\times {10}^5\times 2.4\times {10}^{-3}}{8.314\times 295}=0.099\text{ moles}\text{/min}$

The total 90 moles in the cylinder and there are 0.099 moles are releasing every minute. So, the total time that the cylinder last is calculated by following way,

  $\begin{array}{l}t\text{ = }\frac{\text{Total moles}}{\text{Number of moles per minute}}\\ t=\frac{90}{0.099}=909\text{min}\end{array}$

Conclusion:

Thus, the time up to which the cylinder last for a given flow rate is $909\min$ .