What mass of carbon dioxide is present in 1.00 m3 of dry air at a temperature of 21 ∘C and a pressure of 649 torr ?

Question

Cameron Andrews · Accepted Answer

I got #"0.545 g CO"_2(g)#, assuming it is ideal...

If we assume that #"CO"_2# is an ideal gas at this temperature and pressure, then the mols of it are given in the ideal gas law:

#PV = nRT#

Convert these to the proper units...

#P = 649 cancel"torr" xx "1 atm"/(760 cancel"torr") = "0.8539 atm"#

#V = 1.00 cancel("m"^3) xx ((100 cancel"cm")/cancel"1 m")^3 xx cancel"1 mL"/cancel("1 cm"^3) xx "1 L"/(1000 cancel"mL")#

#= 1.00 xx 10^3 "L"#

#T = 21 + "273.15 K" = "294.15 K"#

Therefore, the mols of dry air in general, if it is truly ideal, is...

#n = (PV)/(RT)#

#= ("0.8539 atm"cdot1.00 xx 10^3 "L")/("0.082057 L"cdot"atm/mol"cdot"K"cdot"294.15 K")#

#=# #"35.4 mols dry air"#

In dry air, there is a mol fraction of #chi_(CO_2) = 0.0350%#, or #0.000350#. Thus, the mols of #"CO"_2# in dry air is simply

#"35.4 mols dry air" xx 0.000350 = ul("0.00124 mols CO"_2(g))#

and this has a mass of

#color(blue)(m_(CO_2)) = 0.00124 cancel("mols CO"_2) xx ("44.009 g CO"_2)/cancel"1 mol"#

#=# #color(blue)ul("0.545 g CO"_2)#

Serenity Allen · Answer

undefined To find the mass of carbon dioxide present in 1.00 m^3 of dry air at a temperature of 21°C and a pressure of 649 torr, we can use the ideal gas law and the partial pressure of carbon dioxide in dry air.

First, we need to convert the temperature from Celsius to Kelvin using the formula:
$$T(K) = T(°C) + 273.15$$

$$T = 21°C + 273.15 = 294.15 K$$

Next, we convert the pressure from torr to atm since the ideal gas law requires pressure in atmospheres:
$$P(atm) = \frac{649\, \text{torr}}{760\, \text{torr/atm}} = 0.854\, \text{atm}$$

Now, we can use the ideal gas law equation:
$$PV = nRT$$

Where:
- $P$ is the pressure in atm
- $V$ is the volume in liters
- $n$ is the number of moles
- $R$ is the ideal gas constant (0.0821 L atm/mol K)
- $T$ is the temperature in Kelvin

Rearranging the equation to solve for moles ($n$):
$$n = \frac{PV}{RT}$$

We know that dry air is composed of approximately 0.04% carbon dioxide by volume. So, to find the moles of carbon dioxide, we multiply the total volume of air by the fraction of carbon dioxide:
$$n_{CO2} = (1.00\, \text{m}^3)(0.0004)$$

Now, we can use the mass formula:
$$m = n \times M$$

Where:
- $m$ is the mass of carbon dioxide
- $n$ is the number of moles of carbon dioxide
- $M$ is the molar mass of carbon dioxide (44.01 g/mol)

$$m = n_{CO2} \times M = (1.00\, \text{m}^3)(0.0004)(44.01\, \text{g/mol})$$