Question

A $6.0cm-\mathrm{diameter},10cm-\mathrm{long}$cylinder contains $100mg$of oxygen (O₂) at a pressure less than $1atm$. The cap on one end of the cylinder is held in place only by the pressure of the air. One day when the atmospheric pressure is $100kPa$, it takes a $184N$force to pull the cap off. What is the temperature of the gas?

Short answer

The temperature of the gas is$380K$

Step-by-step solution

Defining Ideal Gas law

Given : A $6.0-cm$-diameter, $10-cm$-long cylinder contains $100mg$of oxygen at a pressure less than $1atm$. When atmospheric pressure is $100kPa$, it takes a $184N$force to pull the cap off.

The ideal gas law ($PV\mathit{}\mathit{=}\mathit{}nRT$) relates the macroscopic properties of ideal gases.

Since the pressure$p$ comes from inside the gas, we can apply the ideal gas law to find an expression for the temperature
$$\begin{gathered} PV=nRT \\ \Rightarrow T=\frac{PV}{nR} \end{gathered}$$
Knowing that we are giving the mass and not the number of moles, we can rewrite the previous formula as follows.
$T=\frac{pMV}{mR}$, where $M$is the molar mass of the gas inside and $m$is the mass of that gas.

Finding the temperature of the gas 

Based onNewton's first law and the definition of pressure, we can write for the cap that
$$\begin{gathered} pA+F=p_{a}A \\ \Rightarrow p=p_a-\frac{F}{A} \end{gathered}$$
Where $p$is the internal pressure, $p_a$is the atmospheric pressure,

$A=\pi R²=\frac{\pi D²}{4}$is the area of the cap and$F$is the force we have to apply to move the cap.
We can now substitute the expression we found for pressure into our result for temperature:

$T=\frac{pMV}{mR}=\frac{(p_a-{\displaystyle \frac{F}{A}})MV}{mR}$

Remembering that for a cylinder$V=Ah$, we can further simplify to If we apply our numerical input data, we find the value for the temperature inside as follows.
$$\begin{gathered} T=\frac{(p_a-{\displaystyle \frac{F}{A}})MV}{mR}=\frac{(p_a-{\displaystyle \frac{F}{A}})MAh}{mR} \\ =\frac{(A{p}_a-{\displaystyle F})Mh}{mR} \\ \Rightarrow T=\frac{(\frac{\pi D²}{4}{p}_a-F)Mh}{mR} \\ =\frac{Mh(\pi D²p_a-4F)}{4mR} \end{gathered}$$

Inserting the values, we get$\mathit{T}\mathbf{=}\mathbf{}\mathbf{380}\mathbf{}\mathit{k}$

$T=\mathbf{380}\mathbf{}\mathit{K}$