Solved Problem on Gases

The fire extinguishers sold for automobiles are in the shape of the cylindrical capsule, with hemispherical ends, as shown in the figure. They are made of iron and contain about 1 liter of CO₂, under a pressure of 2.8 atmospheres at a temperature of 21°C. Assume that CO₂ behaves as an ideal gas.
a) Calculate the volume of iron used in making the capsule in cm³;
b) Calculate the CO₂ pressure, in atmospheres, at 0°C.

Problem data:

Length of the cylindrical part of the extinguisher: _L = 28 cm;
Diameter of cylinder and hemispheres: d = 8 cm;
Volume of CO₂ contained in the extinguisher: V_g = 1 ℓ.

Initial State	Final State
t_i = 21ºC;	t_f = 0ºC
p_i = 2.8 atm	p_f

Solution

First, we must convert the units of the problem, the volume of the gas given in liters (ℓ) to cubic centimeters (cm³), the temperature given in degrees Celsius (°C) to kelvins (K)

\[ \begin{gather} V_{g}=1\;\cancel{\ell}\times\frac{1000\;\text{cm}^{3}}{1\;\cancel{\ell}}=1000\;\text{cm}^{3} \end{gather} \]

For temperature, we have the conversion formula from degrees Celsius to kelvins

\[ \begin{gather} \bbox[#99CCFF,10px] {T=t_{c}+273} \end{gather} \]

\[ \begin{gather} T_{1}=21+273=294\;\text{K}\\[10pt] T_{2}=0+273=273\;\text{K} \end{gather} \]

a) The volume of iron used to build the extinguisher will be obtained by the volume of the capsule minus the volume of the gas contained in it. To calculate the volume of the capsule, divide it into three parts by separating the hemispheres from the ends of the central cylinder (Figure 1). Uniting the two hemispheres, we have a sphere, calculating the volume of this, and adding it to the volume of the cylinder, we will have the volume of the capsule.

Figure 1

The radii of the sphere and cylinder will be half the diameter of the capsule given in the problem

\[ \begin{gather} r=\frac{d}{2}\\[5pt] r=\frac{8}{2}\\[5pt] r=4\;\text{cm} \end{gather} \]

The volume of a sphere is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {V=\frac{4}{3}\pi r^{3}} \end{gather} \]

We use π = 3.14, so the volume of the sphere will be

\[ \begin{gather} V_{1}=\frac{4}{3}\times 3.14\times 4^{3}\\[5pt] V_{1}=\frac{4}{3}\times 3.14\times 64\\[5pt] V_{1}\approx 267.9\;\text{cm}^{3} \tag{I} \end{gather} \]

The volume of a cylinder is given by

\[ \begin{gather} \bbox[#99CCFF,10px] {V=Bh} \tag{II} \end{gather} \]

where B is the area of the base of the cylinder, given by the area of a circle

\[ \begin{gather} \bbox[#99CCFF,10px] {B=\pi r^{2}} \tag{III} \end{gather} \]

substituting expression (III) into expression (II), the volume of the cylinder will be

\[ \begin{gather} V=\pi r^{2}h \end{gather} \]

where h = 28 cm the height of the cylinder, we will have the volume

\[ \begin{gather} V_{2}=3.14\times 4^{2}\times 28\\[5pt] V_{2}=1406.7\;\text{cm}^{3} \tag{IV} \end{gather} \]

Then the volume of the capsule will be the sum of the results (I) and (IV)

\[ \begin{gather} V_{c}=V_{1}+V_{2}\\[5pt] V_{c}=267.9+1406.7\\[5pt] V_{c}=1674.6\;\text{cm}^{3} \end{gather} \]

The volume of iron will be the difference between the volume of the capsule and the volume occupied by the gas

\[ \begin{gather} V_{f}=V_{c}-V_{g}\\[5pt] V_{f}=1674.6-1000 \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {V_{f}=674.6\;\text{cm}^{3}} \end{gather} \]

b) During the change in temperature from 21ºC to 0ºC the volume does not change (it remains the volume of the extinguisher), we have an isovolumetric (or isochoric or isometric) transformation, for the calculation of the final pressure we use the Gay-Lussac's Law

\[ \begin{gather} \bbox[#99CCFF,10px] {\frac{p_{i}}{T_{i}}=\frac{p_{f}}{T_{f}}} \end{gather} \]

\[ \begin{gather} \frac{2.8}{294}=\frac{p_{f}}{273}\\[5pt] p_{f}=\frac{2.8}{294}\times 273 \end{gather} \]

\[ \begin{gather} \bbox[#FFCCCC,10px] {p_{f}=2.6\;\text{atm}} \end{gather} \]

Fisicaexe - Physics Solved Problems by Elcio Brandani Mondadori is licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 4.0 International License .