Baseball reporters say that long fly balls that would have carried for home runs in July "die" in the cool air of Octo - ber and are caught. The idea behind this observation is that a baseball carries better when the air is less dense. Dry air is a mixture of gases with an effective molar mass of 29.0 g mol21. (a) Compute the density of dry air on a July day when the temperature is 95.0°F and the pressure is 1.00 atm. (b) Compute the density of dry air on an October evening when the temperature is 50.0°F and the pressure is 1.00 atm. (c) Suppose that the humidity on the July day is 100%; thus, the air is saturated with water vapor. Is the density of this hot, moist air less than, equal to, or greater than the density of the hot, dry air computed in part (a) ? In other terms, does high humidity favor the home run?

Question

Keenan

Physics

21 June, 22:45

Baseball reporters say that long fly balls that would have carried for home runs in July "die" in the cool air of Octo - ber and are caught. The idea behind this observation is that a baseball carries better when the air is less dense. Dry air is a mixture of gases with an effective molar mass of 29.0 g mol21. (a) Compute the density of dry air on a July day when the temperature is 95.0°F and the pressure is 1.00 atm. (b) Compute the density of dry air on an October evening when the temperature is 50.0°F and the pressure is 1.00 atm. (c) Suppose that the humidity on the July day is 100%; thus, the air is saturated with water vapor. Is the density of this hot, moist air less than, equal to, or greater than the density of the hot, dry air computed in part (a) ? In other terms, does high humidity favor the home run?

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Answers (1)

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Broderick Crane · Answer 1

Consider the followig calculation

Explanation:

a) use deal equation:

PV = nRT

ρ = m/V, = = => V = m/ρ

therefore,

ρ = Pm/RT

convert 95 oF in degree

95 oF = 308.15 K

1 atm = 1.013 * 105 pascal

ρ = 1.013*105 * 29 * / 8.314 * 308.15

= 1.146 kg/m3

b) again use ideal gas equation:

ρ = Pm/RT

T = 50 oF = 283.15 K

1 atm = 1.013 * 105 pascal

molar mass will be same

ρ = 1.013 * 105 * 29 / 8.314 * 283.15

ρ = 1.248 kg / m3

So,

c). more than density of the hot, dry air computed in part (a)