Chemistry High School

A container in the shape of a cube 10.0 cm on each edge contains air (with equivalent molar mass 28.9 g/mol ) at atmospheric pressure and temperature 300 K. Find: (a) the mass of the gas, (b) the gravitational force exerted on it, and (c) the force it exerts on each face of the cube. (d) Why does such a small sample exert such a great force?

Answers

Answer 1

(a) The mass of the gas is 1.16 g.

(b) The gravitational force exerted on it is 11.4 N.

(c) The force it exerts on each face of the cube is 998 N.

(d) The reason why such a small sample exerts such a great force is that the collisions of the gas molecules with the walls of the container can produce a large force because the gas molecules are moving very rapidly and colliding with the walls many times per second.

(a) The volume of the container is V = 10.0 cm × 10.0 cm × 10.0 cm = 1000 cm³ = 1.00 L. The mass of the gas can be calculated by the ideal gas law, which states:

P V = n R T

where P is the pressure, V is the volume, n is the number of moles of gas, R is the ideal gas constant, and T is the temperature. Rearranging this equation to solve for n, the number of moles of gas:

n = P V / (R T)

The ideal gas constant R = 8.31 J/(mol·K). Substituting in the given values for P, V, and T:

n = (101,325 Pa)(0.00100 m³) / [(8.31 J/(mol·K))(300 K)] = 0.0402 mol

The mass of the gas can be calculated using the molar mass and the number of moles:

m = n M

where M is the molar mass. Substituting in the given values:

m = (0.0402 mol)(28.9 g/mol) = 1.16 g

(b) The gravitational force exerted on the gas is given by:

F = m g

where g is the acceleration due to gravity. Substituting in the given values:

F = (1.16 g)(9.81 m/s²) = 11.4 N

(c) The force exerted on each face of the cube is equal and opposite to the pressure of the gas on the walls of the container. The pressure of an ideal gas is given by:

P = n R T / V

Substituting in the given values:

P = (0.0402 mol)(8.31 J/(mol·K))(300 K) / (0.00100 m³) = 99,800 Pa

The force on each face of the cube is given by:

F = P A

where A is the area of one face of the cube. Substituting in the given values:

F = (99,800 Pa)(0.100 m × 0.100 m) = 998 N

(d) The force exerted by the gas on each face of the cube is due to the pressure of the gas. The pressure is caused by the collisions of the gas molecules with the walls of the container. Even though the mass of the gas is small, the collisions of the gas molecules with the walls of the container can produce a large force because the gas molecules are moving very rapidly and colliding with the walls many times per second.

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Related Questions

What is the Phase constant?

Express your answer in radians to three significant figures.

I know the phase constant is 3pi/2 but I don't how to convert it to three sig figs. Please help!

Answers

The phase constant, expressed in radians to three significant figures, is approximately 4.71 rad.

To convert the phase constant, which is given as 3π/2, to three significant figures, we need to evaluate the numerical value of the expression.

The value of π (pi) is approximately 3.14159, and dividing 3 by π gives us 0.95493. Multiplying this value by 2, we get 1.90987. To achieve three significant figures, we round this value to 1.91.

Hence, the phase constant, 3π/2, can be approximated as 1.91.

It's important to note that rounding the numerical value of the expression to three significant figures does not affect the symbolic representation, which remains 3π/2. However, when expressing the value in numerical form, rounding to three significant figures provides a more concise and accurate representation.

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