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mote1985 [20]
2 years ago
8

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. Fin

d the force exerted by the gas on one of the walls of the container in kN.
Physics
1 answer:
rodikova [14]2 years ago
5 0

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)\\

<u>F = 74.08 KN</u>

<u></u>

Learn more about the general gas equation here:

brainly.com/question/24645007?referrer=searchResults

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3 years ago
Two particles, with charges of 20.0 nC and -20.0 nC, are fixed at points with coordinates &lt;0, 4.00 cm&gt; and &lt;0, -4.00 cm
Dmitry [639]

Answer:

Explanation:

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=  9 x 10⁹ x [  q₁q₂ / r₁₂ +  q₂q₃ / r₂₃ +  q₁q₃ / r₁₃ ]

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r₁₂ = 8 cm , r₂₃ = 4 cm , r₁₃ = 4 cm.

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b)

Potential at the point of fourth charge due to three charges of 20 nC , - 20 nC and 10 nC at the centre

9 x 10⁹ [ 20 x 10⁻⁹ / .05 + - 20 x 10⁻⁹ / .05 + 10 x 10⁻⁹ / .03 ]

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