Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store
seo-qna
SearchIcon
banner

Pressure of a mixture of $4\,g$ ${{O}_{2}}$and $2\,g$of ${{H}_{2}}$confined in a bulb of $1\,litre$at ${{0}^{{\mathrm O}}}C$is
A.$25.215\,\,atm$ [AIIMS$2000$]
B.$31.205\,\,atm$
C.$45.215\,\,atm$
D.$15.210\,\,atm$

Answer
VerifiedVerified
164.7k+ views
Hint: Perfect gas law describes the relationship between the volume, pressure, and temperature of a gas in the limit of high temperature and low pressure implying that the gas molecules move almost independently of each other. By putting the given values in the ideal gas equation we can easily calculate pressure.

Formula Used:Perfect gas law or ideal gas law equation is $PV=nRT$
Or,$P=\dfrac{nRT}{V}$
Here $P=$Pressure of gas molecules
$R=$Universal gas constant
$T=$Temperature
$V=$Volume of gas molecules
$n=$Mole number of gaseous molecules
Again,$n=\dfrac{W}{M}$
Where $W$and $M$denotes the weight and molar mass of the gas molecule

Complete step by step solution:Mole number of ${{H}_{2}}$molecules,${{n}_{{{H}_{2}}}}=\dfrac{{{W}_{{{H}_{2}}}}}{{{M}_{{{H}_{2}}}}}$
Given, ${{W}_{{{H}_{2}}}}=2\,g$, ${{M}_{{{H}_{2}}}}=2\,g/mol$
$\therefore {{n}_{{{H}_{2}}}}=\dfrac{2\,g}{2\,g/mol}=1\,mol$
Hence the total number of moles,$n={{n}_{{{O}_{2}}}}+{{n}_{{{H}_{2}}}}=(0.125+1)\,mol=1.125\,mol$
Volume,$v=1\,litre$ , Temperature,$T={{0}^{O}}C=273K$and Universal gas constant,$R=0.082\,L\,atm\,/mol/K$
Putting these values in equation (i) we get,
$P=\dfrac{nRT}{V}$
Or,$P=\dfrac{1.125\times 0.082\times 273K}{1}=25.18\,atm$
Therefore pressure$25.18\,atm$ nearly equals to $25.215\,atm$. Generally, the ideal gas law equation is the hypothetical equation. As an ideal gas is a hypothetical entity that does not exist in real life. The main purpose of creating the ideal gas law was to express the relationship between pressure, volume, number of moles of gases, and temperature.
The equation of ideal gas is, $PV=nRT$
Or,$P=\dfrac{nRT}{V}$ ……….(i)
Mole number of ${{O}_{2}}$molecules,${{n}_{{{O}_{2}}}}=\dfrac{{{W}_{{{O}_{2}}}}}{{{M}_{{{O}_{2}}}}}$
Given ${{W}_{{{O}_{2}}}}=4\,g$, ${{M}_{{{O}_{2}}}}=32\,g/mol$
$\therefore {{n}_{{{O}_{2}}}}=\dfrac{4\,g}{32\,g/mol}=0.125\,mol$

Thus, option (A) is correct.

Note: Ideal gas law equation is only valid at low pressure and high temperature. But gas molecules deviate from ideal behavior at high pressure and low temperature since ideal gas molecules are slow-moving, closely packed molecules are more likely to interact with each other.