Question

The volume of $1\,\,g$each of methane $(C{{H}_{4}})$, ethane $({{C}_{2}}{{H}_{6}})$, propane $({{C}_{3}}{{H}_{8}})$, and butane$({{C}_{4}}{{H}_{10}})$was measured at $350\,K$ and $1\,\,atm$. What is the volume of butane [NCERT $1981$]
A.$495\,\,c{{m}^{3}}$
B.$600\,\,c{{m}^{3}}$
C.$900\,\,c{{m}^{3}}$
D.$1700\,\,c{{m}^{3}}$

Answer 1

Hint: In an ideal gas the correlations between pressure, temperature, volume, and quantity of gas can be expressed by the Ideal gas law. Here four different gasses methane, ethane, propane, and butane are mentioned and we can calculate the volume of butane by applying the ideal law equation.

Formula Used:The mathematical expression of the ideal gas law is:
$PV=\dfrac{m}{M}RT$
Here $P\And V$denotes the pressure and volume of the gas.
 $R\And T$represents the Universal gas constant and the temperature.
$m\And M$denotes given mass and molar mass of the gas

Complete step by step solution:A perfect gas or ideal gas is a gas in which the gaseous particles do not repel and attract each other and take up no space. An ideal gas law equation provides a relation between the pressure$(P)$, volume$(V)$, temperature$(T)$, and quantity of gas by the equation:$PV=nRT$
Here $n$is the quantity of gas expressed in the unit of mole.
Further, the number of moles$n$ can be written in terms of the given mass,$m$ and molar mass,$M$ of the constituent in the following way,
$\therefore n=\dfrac{m}{M}$
Putting this equation in the ideal gas equation we get,
$PV=nRT$
Or,$PV=\dfrac{m}{M}RT$
Or,$V=\dfrac{m}{M}\times \dfrac{RT}{P}$ …….(i)
The molar mass of butane,${{M}_{{{C}_{4}}{{H}_{10}}}}=$($4\times $Atomic weight of Carbon)$+$($10\times $Atomic weight of Hydrogen)
$\therefore {{M}_{{{C}_{4}}{{H}_{10}}}}=(4\times 12+10\times 1)=58g/mol$
Given the mass of butane,$m=1g$
$P=1\,atm$
$T=350K$
And $R=0.082\,\,L\,atm\,/mol/K$
Putting these values in equation (i),
$V=\dfrac{1g}{58\,g/mol}\times \dfrac{0.082\,L\,atm/mol/K\times 350K}{1atm}$
Or,$V=0.495\,L$
Or,$V=495\,c{{m}^{3}}$ [since $1L=1d{{m}^{3}}$;$1dm=10cm$]
Therefore the volume of butane is $495c{{m}^{3}}$.

Thus option (A) is correct.

Note: Unit conversions are important in all science subjects although they may seem more important in chemistry as many calculations use different units of measurement. This is crucial in order to have accuracy and avoid confusion in measurement.