Chapter 18: Question 46 (page 512)

A $6.0 c m - diameter, 10 c m - long$ cylinder contains $100 m g$ of oxygen (O₂) at a pressure less than $1 a t m$ . 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 $100 k P a$ , it takes a $184 N$ force to pull the cap off. What is the temperature of the gas?

Short Answer

Expert verified

The temperature of the gas is $380 K$

Step by step solution

Defining Ideal Gas law

Given : A $6.0 - c m$ -diameter, $10 - c m$ -long cylinder contains $100 m g$ of oxygen at a pressure less than $1 a t m$ . When atmospheric pressure is $100 k P a$ , it takes a $184 N$ force to pull the cap off.

The ideal gas law ( $P V = n R T$ ) 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
$P V = n R T \Rightarrow T = \frac{P V}{n R}$
Knowing that we are giving the mass and not the number of moles, we can rewrite the previous formula as follows.
$T = \frac{p M V}{m R}$ , 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
$p A + F = p_{a} A \Rightarrow p = p_{a} - \frac{F}{A}$
Where $p$ is the internal pressure, $p_{a}$ is the atmospheric pressure,

$A = π R ² = \frac{π D ²}{4}$ 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:

$T = \frac{p M V}{m R} = \frac{(p_{a} - \frac{F}{A}) M V}{m R}$

Remembering that for a cylinder $V = A h$ , we can further simplify to If we apply our numerical input data, we find the value for the temperature inside as follows.
$T = \frac{(p_{a} - \frac{F}{A}) M V}{m R} = \frac{(p_{a} - \frac{F}{A}) M A h}{m R} = \frac{(A p_{a} - F) M h}{m R} \Rightarrow T = \frac{(\frac{π D ²}{4} p_{a} - F) M h}{m R} = \frac{M h (π D ² p_{a} - 4 F)}{4 m R}$