Question

How would you find the number of grams of $C{{O}_{2}}$ that exert a pressure of 785 torrs at a volume of 32.5 L and a temperature of $32{}^\circ C$.

Answer 1

A hypothetical observation of an ideal gas is taken that obey Boyle’s law, Charle’s law and Avogadro’s law under similar conditions or temperature and pressure. The equation of ideal gas is also called the equation of state.
Formula used:
Ideal gas equation (equation of state)
$PV=nRT$

Complete answer:
We have been given a gas carbon dioxide that has quantities of pressure, temperature and volume as:
P= 785 torrs
V= 32.5 L
T= $32{}^\circ C$
Now considering this gas to be ideal, the ideal gas equation is used,
$~PV=nRT$
Where, P is the pressure, V is the volume, n is number of moles, T is temperature and R is the gas constant and its value is in accordance with pressure and volume that is $62.364\,L\,torr\,mo{{l}^{-1}}\,{{K}^{-1}}$
So, putting the given values of pressure, temperature, volume and gas constant into the ideal gas equation we get:
$785\,torr\times 32.5L=n\,~62.36\,L\,torr\,mo{{l}^{-1}}\,{{K}^{-1}}\times (32+273)K$
Since, n is unknown so,
$n=\dfrac{785\,torr\,\times 32.5\,L}{62.36\,L\,torr\,mo{{l}^{-1}}\,{{K}^{-1}}\times (32+273)K}$
So, number of moles,
n= 1.34 mol
Now we have to take out the mass of carbon dioxide using number of moles, as we know that,$n=\dfrac{weight}{molar\,mass}$
Here, weight is unknown, molar mass of $C{{O}_{2}}=44\,g\,mo{{l}^{-1}}$ and n= 1.34 mol
Therefore, $1.34\,mol=\dfrac{weight}{44\,g\,mo{{l}^{-1}}}$
Weight of carbon dioxide = $1.34\,mol\times 44\,g\,mo{{l}^{-1}}$
Weight = 58.96 g
Hence, the number of grams of carbon dioxide $C{{O}_{2}}$ is found to be 58.96 g.

Note:
Gas constant has different values in various units of temperature and pressure like, $8.314\,J{{K}^{-1\,}}\,mo{{l}^{-1}}$, $2\,cal\,mo{{l}^{-1}}\,{{K}^{-1}}$, $0.082\,L\,atm\,{{K}^{-1}}\,mo{{l}^{-1}}$.
The value of temperature should be converted into Kelvin as that in the gas constant.