Question: How many grams of fluorine gas will exert a pressure of 134 atm in a 3.2 liter container at \({{40}^...

Question

How many grams of fluorine gas will exert a pressure of 134 atm in a 3.2 liter container at ${{40}^{\circ }}C$ ?

Answer 1

Solution

To solve the equation, we use the ideal gas equation and the equation is as follows: $PV=nRT$ , where P is pressure of the gas, V is volume of the gas, R is gas constant, n is number of moles of gas and T is the temperature.

Complete answer:
In the question it is asked how many grams of fluorine gas will exert the pressure of 134atm in a volume of 3.2liter container if the temperature value of the container is at ${{40}^{\circ }}C$ .
So we have to find the amount of fluorine gas in grams.
We know that several gas laws are formulated to explain the properties of gas in terms of volume of the gas, absolute temperature of the gas, the number of moles of gas or amount of gaseous sample and the pressure exerted on the container by the gas molecules etc.
There are four main laws: Avogadro law, Boyle's law, Charles's aw and Gay-Lussac law
The combination of these four laws will give the ideal gas equation.
The ideal gas equation is as follows: $PV=nRT$
From the ideal equation we can calculate the parameters like pressure, volume, temperature and number of moles of gas or amount of gas etc.
Now let us solve the given question using this equation. Let us first write down the values of variables given.
The given data is:
$P=134atm$
$V=3.2L$
$T={{40}^{\circ }}C=273+40=313K$
The value of R will varies with the unit here the value of R is, $0.0821L.atm/K.mol$
Substitute these values in the ideal gas equation, we will get a number of moles of gas.
$PV=nRT$
$n=\dfrac{PV}{RT}$
$n=\dfrac{134\times 3.2}{0.0821\times 313}=16.7mol$
Now we got the number of moles of gas. From the number of moles we have to find the mass of gas.
We know the formulae relating the number of moles and mass.
$No.\,of\,moles=\dfrac{given\,mass}{Molecular\,mass}$
$mass=Molecular\,mass\times no.\,of\,moles$
$mass=16.7\times 38=634.6g$

Note:
We took the mass value of fluorine gas as $38g/mol$ instead of $19g/mol$ , because fluorine exists as molecule and the gaseous sample of the gas contains fluorine molecule, so we should add up the mass of two F atoms or multiply the mass of an F atom with 2.
We could find other parameters like pressure, temperature etc. with respect to the data given.