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Question: The molecular mass of \[{O_2}\] and \[S{O_2}\] are 32 and 64 respectively. If one litre of \[{O_2}\]...

The molecular mass of O2{O_2} and SO2S{O_2} are 32 and 64 respectively. If one litre of O2{O_2} at 150C{15^0}C and 750mm750mm pressure contains NN molecules, the number of molecules in two litre of SO2S{O_2} under the same conditions of temperature and pressure will be:

A. 2N B. N C. N2 D. 4N  A.{\text{ }}2N \\\ B.{\text{ }}N \\\ C.{\text{ }}\dfrac{N}{2} \\\ D.{\text{ }}4N \\\
Explanation

Solution

Hint: In order to solve the given question first we will consider the number of molecules in both of the compounds in terms of some unknown variable, then we will use the ideal gas equation individually for both of the cases in the problem to find the unknown variables. Further we will divide the equations to find the relationship between the molecules.

Complete step by step solution: Formula used- PV=nRT,Nm=NA×nPV = nRT,{N_m} = {N_A} \times n
As we know that the ideal gas equation is given as:
PV=nRTPV = nRT
Where
P is the pressure of the gas
V is the volume of the gas
n is the number of moles of the gas
R is the gas constant
T is the temperature of the gas.
Given that the number of molecules of oxygen is N
Let us consider the number of molecules of sulphur dioxide as y.

Given that:
For oxygen or O2{O_2}
We have:
Pressure =750mm = 750mm
Volume = 1 litre
Temperature =150C = {15^0}C
Let R is the gas constant and n is the number of moles of the gas.
So, let us use the ideal gas equation to find the number of moles.
Substituting the values in the ideal gas equation we get:
PV=nRT n=PVRT n=750×1R×15(L mm (0C)1) n=50R(L mm (0C)1)  \because PV = nRT \\\ \Rightarrow n = \dfrac{{PV}}{{RT}} \\\ \Rightarrow n = \dfrac{{750 \times 1}}{{R \times 15}}\left( {L{\text{ }}mm{\text{ }}{{\left( {^0C} \right)}^{ - 1}}} \right) \\\ \Rightarrow n = \dfrac{{50}}{R}\left( {L{\text{ }}mm{\text{ }}{{\left( {^0C} \right)}^{ - 1}}} \right) \\\
Now as we have the number of moles of the gas, let us find the number of molecules in the gas.
We know that the relation between number of moles and number of molecules is given as:
Nm=NA×n{N_m} = {N_A} \times n
Where,
Nm{N_m} represents the number of molecules.
NA{N_A} represents Avagadro’s number.
And n represents the number of moles.
Let us use the above formula to find the number of molecules N of oxygen.
Nm=NA×n N=NA×50R..........(1)  \because {N_m} = {N_A} \times n \\\ \Rightarrow N = {N_A} \times \dfrac{{50}}{R}..........(1) \\\
Similarly let us find the number of molecules of sulphur dioxide.
For sulphur dioxide or SO2S{O_2}
We have:
Pressure =750mm = 750mm
Volume = 2 litre
Temperature =150C = {15^0}C
Let R is the gas constant and n is the number of moles of the gas.
So, let us use the ideal gas equation to find the number of moles.
Substituting the values in the ideal gas equation we get:
PV=nRT n=PVRT n=750×2R×15(L mm (0C)1) n=100R(L mm (0C)1)  \because PV = nRT \\\ \Rightarrow n = \dfrac{{PV}}{{RT}} \\\ \Rightarrow n = \dfrac{{750 \times 2}}{{R \times 15}}\left( {L{\text{ }}mm{\text{ }}{{\left( {^0C} \right)}^{ - 1}}} \right) \\\ \Rightarrow n = \dfrac{{100}}{R}\left( {L{\text{ }}mm{\text{ }}{{\left( {^0C} \right)}^{ - 1}}} \right) \\\
Now as we have the number of moles of the gas, let us find the number of molecules in the gas.
We know that the relation between number of moles and number of molecules is given as:
Nm=NA×n{N_m} = {N_A} \times n
Where,
Nm{N_m} represents the number of molecules.
NA{N_A} represents Avagadro’s number.
And n represents the number of moles.
Let us use the above formula to find the number of molecules y of sulphur dioxide.
Nm=NA×n y=NA×100R.........(2)  \because {N_m} = {N_A} \times n \\\ \Rightarrow y = {N_A} \times \dfrac{{100}}{R}.........(2) \\\
Now since we have the number of molecules of both oxygen and sulphur dioxide let us divide equation (2) by equation (1) in order to find the relation between the number of molecules of oxygen and sulphur dioxide.
y=NA×100R&N=NA×50R yN=(NA×100R)(NA×50R)  \because y = {N_A} \times \dfrac{{100}}{R}\& N = {N_A} \times \dfrac{{50}}{R} \\\ \Rightarrow \dfrac{y}{N} = \dfrac{{\left( {{N_A} \times \dfrac{{100}}{R}} \right)}}{{\left( {{N_A} \times \dfrac{{50}}{R}} \right)}} \\\
Now let us further solve the above equation to find the relation.
yN=NA×100NA×50×RR yN=2 y=2N  \Rightarrow \dfrac{y}{N} = \dfrac{{{N_A} \times 100}}{{{N_A} \times 50}} \times \dfrac{R}{R} \\\ \Rightarrow \dfrac{y}{N} = 2 \\\ \Rightarrow y = 2N \\\
Hence, the number of molecules of sulphur dioxide under the same conditions are 2N.

So, the correct answer is option A.

Note: The given problem can also be solved directly by the help of Avogadro’s law by directly substituting the values in the formula and finding the relation. But the method used above is the basic one and easy to understand. Students must remember that all the laws such as Avogadro’s law and other laws are derived from the ideal gas equation only after taking in some particular conditions. So ideal gas law is the most important amongst all and must be remembered by the students.