The equation to be used are:
PM = ρRT
PV = nRT
where
P is pressure, M is molar mass, ρ is density, R is universal gas constant (8.314 J/mol·K), T is absolute temperature, V is volume and n is number of moles
The density of air at 23.5°C, from literature, is 1.19035 kg/m³. Its molar mass is 0.029 kg/mol.
PM = ρRT
P(0.029 kg/mol) = (1.19035 kg/m³)(8.314 J/mol·K)(23.5+273 K)
P = 101,183.9 Pa
n = 0.587 g * 1 kg/1000 g * 1 mol/0.029 kg = 0.02024 mol
(101,183.9 Pa)V = (0.02024 mol)(8.314 J/mol·K)(23.5+273 K)
Solving for V,
V = 4.931×10⁻⁴ m³
Since 1 m³ = 1000 L
V = 4.931×10⁻⁴ m³ * 1000
V = 0.493 L
Answer :
The pressure of carbon dioxide in the atmosphere in mmHg is 0.239mmHg.
The pressure of carbon dioxide in the atmosphere in atm is
.
Explanation :
The conversion used for pressure from torr to mmHg is:
1 torr = 1 mmHg
The conversion used pressure from torr to atm is:
1 atm = 760 torr
or,

As we are given the pressure of carbon dioxide in the atmosphere 0.239 torr. Now we have to determine the pressure of carbon dioxide in the atmosphere in mmHg and atm.
<u>Pressure in mmHg :</u>
As, 
So, 
Thus, the pressure of carbon dioxide in the atmosphere in mmHg is 0.239mmHg.
<u>Pressure in atm:</u>
As, 
So, 
Thus, the pressure of carbon dioxide in the atmosphere in atm is
.
Stir it,
Or as warmer water makes solutes dissolve faster Sarah can do that