Forty miles above the earth's surface the temperature is 250 K and the pressure is only 0.20 mm Hg. What is the density of air (in grams per liter) at this altitude? (Assume the molar mass of air is 28.96 g/mol.)

Question

Forty miles above the earth's surface the temperature is 250 K and the pressure is only 0.20 mm Hg. What is the density of air (in grams per liter) at this altitude? (Assume the molar mass of air is 28.96 g/mol.)

PV=nRT
n/V= molar desity= P/RT
mass density= molar density*molmass
= P*(molmass)/RT

Watch your units.

Answer 1

0.20 mm Hg = 0.20*133.322 Pa

mass density= (0.20*133.322*28.96)/(8.314*250)

= 0.0014 g/L

Answer 2

To find the density of air at this altitude, we can use the ideal gas law equation PV = nRT where:

P = pressure (0.20 mm Hg)
V = volume (1 liter)
n = number of moles of air (unknown)
R = gas constant (0.0821 L·atm/(mol·K))
T = temperature (250 K)
The molar mass of air (M) is given as 28.96 g/mol.

First, let's convert the pressure from mm Hg to atm:
1 atm = 760 mm Hg
0.20 mm Hg = 0.20/760 atm

Now we can rearrange the equation PV = nRT to solve for the number of moles (n):
n = PV / RT
= (0.20/760 atm) * (1 liter) / (0.0821 L·atm/(mol·K)) * (250 K)

n ≈ 0.00149 moles

Next, we can calculate the molar density (n/V):
molar density = n / V
= 0.00149 moles / 1 liter

Now, let's calculate the mass density by multiplying the molar density by the molar mass of air:
mass density = molar density * molar mass
= (0.00149 moles / 1 liter) * (28.96 g/mol)

Finally, we have the density of air at this altitude:
mass density ≈ 0.043 g/L

Therefore, the density of air at this altitude is approximately 0.043 grams per liter.

Answer 3

To find the density of air at this altitude, we can use the ideal gas law equation PV = nRT. Since we want the density in grams per liter, we need to manipulate this equation.

First, let's convert the pressure from mm Hg to atm, since the ideal gas constant R has units of atm·L/mol·K. We can do this by dividing the pressure by 760 (since 760 mm Hg = 1 atm).

Pressure = 0.20 mm Hg / 760 mm Hg/atm ≈ 0.00026316 atm

Next, we can rearrange the equation to solve for n/V, which represents the molar density:

n/V = P / RT

Now, substitute the values we have:

n/V = 0.00026316 atm / (R * 250 K)

The molar mass of air is given as 28.96 g/mol.

Now, we can calculate the mass density:

mass density = (n/V) * (molar mass)

mass density = (0.00026316 atm / (R * 250 K)) * 28.96 g/mol

Finally, we need to convert this mass density to grams per liter. Since 1 mole of a gas at STP occupies 22.4 liters, we can use this conversion factor:

grams per liter = (mass density) * (22.4 L/mol)

Substitute the value of the mass density we obtained earlier into this equation to find the final answer.

Remember to pay attention to units throughout the calculations to ensure consistency.