A sealed cubical container 6.2 cm on a side contains three times Avogadro's number of molecules at a temperature of 32°C. Find the force (in kN) exerted by the gas on one of the walls of the container. (Answer to 1 decimal place)

Question

02-11-2023
Physics

High School

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A sealed cubical container 6.2 cm on a side contains three times Avogadro's number of molecules at a temperature of 32°C. Find the force (in kN) exerted by the gas on one of the walls of the container. (Answer to 1 decimal place)

Answer :

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rathipushpendra345 rathipushpendra345 · Answer 1 · 2024-04-17T19:16:30-04:00

The force exerted by the gas on one of the walls of the container is approximately 4.3 kN.

To calculate the force, we can use the ideal gas law, which states that PV = nRT, where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature in Kelvin. First, we need to determine the number of moles of gas in the container. Since the container contains three times Avogadro's number of molecules, the number of moles (n) is 3 times Avogadro's number divided by Avogadro's number, which simplifies to 3 moles.

Next, we need to convert the temperature to Kelvin. 32°C is equivalent to 305.15 K.

Now, we have the number of moles (n = 3), the temperature (T = 305.15 K), and the volume (V = 0.062 m³) since the container is cubical and each side is 6.2 cm. We can use these values to solve for the pressure (P) using the ideal gas law equation.

Once we have the pressure, we can calculate the force exerted on one of the walls of the container using the formula F = PA, where A is the area of one wall. Since the container is cubical, each wall has an area of [tex](6.2 cm)^2[/tex] = 0.03844 m².

Plugging in the values, we find F = P * A = (3 * 8.314 J/mol·K * 305.15 K) / 0.062 m³ * 0.03844 m² = 4300 N ≈ 4.3 kN.

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