Ideal gasses questions

The 'laws of football' require the ball to have a circumference between 680 mm and 700 mm. The pressure of the air in the ball is required to be between 0.60 x 10⁵ Pa and 1.10 x 10⁵ Pa above atmospheric pressure.

A ball is inflated when the atmospheric pressure is 1.00 x 10⁵ Pa and the temperature is 17 C. When inflated the mass of air inside the ball is 11.4 g and the circumference of the ball is 690 mm. Assume that air behaves as an ideal gas and that the thickness of the material used for the ball is negligible.

Deduce if the inflated ball satisfies the law of football about the pressure.

molar mass of air = 29 g mol^-1

Radius = 690 mm / 6.28 = 110 mm = 0.11m

volume of air = 4/3 × π × 0.11 = 5.55 x 10^-3 m³

T (in kelvin) = 17 + 273 = 290 K

n = 11.4 / 29 = 0.392 mol

Use of pV = nRT = (0.392 × 8.31 × 290) / 5.55 × 10^-3m³

p = 1.70 x 10⁵ Pa

1.70 x 10⁵ - 1.00 × 10⁵ = 0.70 × 10⁵ Pa

Conclusion: this pressure falls within the required band so satisfies the law of football

A cylinder of volume 0.20 m³ contains an ideal gas at a pressure of 130 kPa and a temperature of 290 K. Calculate:

1. the amount of gas, in moles, in the cylinder

2. the average kinetic energy of a molecule of gas in the cylinder

3. the average kinetic energy of the molecules in the cylinder.

1. pV = nRT → 130 x 10³ x 0.20 = n x 8.31 x 290

n = 10.8 mol

2. E_k = 3/2 × kT → E_k = 3 x 1.38 x 10^-23 x 290 = 6.0 × 10^-21 J

3. no. of molecules N = 6.02 x 10²³ x 10.8 = 6.5 x 10²⁴

total k.e. = 6.5 x 10²⁴ x 6.0 x 10^-21 = 3.9 x 10⁴ J

The volume of a fixed mass of an ideal gas is V. The gas exerts pressure p and has thermodynamic temperature T. The temperature of the gas is now increased to 2T. The new pressure exerted by the gas is 3p.

What is the new volume of the gas in terms of V?

2/3 V

Suppose we have four molecules in our gas sample. Their speeds are 3.0, 4.5, 5.2 and 8.3 ms^-1.

Calculate

1. average speed c̅

2. average of the squared speeds c̅²̅

3. square root of c̅²̅ (the r.m.s. speed)

c̅ = (3.0+4.5+5.2+8.3) / 4 = 5.25 ms^-1