Question

A sealed cubical container 10.0 cm on a side contains three times Avogadro's number of molecules at a temperature of 24.0°C. Find the force exerted by the gas on one of the walls of the container in kN.

Answer 1

This question involves the concepts of general gas equation and pressure.

The force exerted by the gas on one of the walls of the container is "74.08 KN".

First, we will use the general gas equation to find out the pressure of the gas:

$PV = nRT$

where,

P = Pressure of the gas = ?

V = Volume of cube = (side length)³ = (10 cm)³ = (0.1 m)³ = 0.001 m³

n = no. of moles = 3 (since molecules equal to avogadro's number make up 1 mole)

R = general gas constant = 8.314 J/mol.K

T = Absolute Temperature = 24°C + 273 = 297 K

Therefore,

$P = \frac{(3)(8.314\ J/mol.k)(297\ K)}{0.001\ m^3}$

P = 7407.78 KPa

Now, the force on one wall can be given as follows:

$P =\frac{F}{A}\\\\F=PA$

where,

A = area of one wall = (side length)² = (0.1 m)² = 0.01 m²

Therefore,

$F=(7407.78\ KPa)(0.01\ m^2)$

F = 74.08 KN

Learn more about the general gas equation here:

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