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Question: When \(0.28litre\) of liquid nitrogen (density = \(0.8gm/mL\) ) is vapourized, what volume does the ...

When 0.28litre0.28litre of liquid nitrogen (density = 0.8gm/mL0.8gm/mL ) is vapourized, what volume does the resulting gas occupy at 27C27^\circ C and 5atm5{\text{atm}} ( Use: R=0.08atm.litre/K.moleR = 0.08{\text{atm}}{\text{.litre/K}}{\text{.mole}} )
A.32.2litre32.2litre
B.38.4 litre38.4{\text{ litre}}
C.41.6 litre41.6{\text{ litre}}
D.57.2 litre57.2{\text{ litre}}

Explanation

Solution

We have to know that an ideal gas is a theoretical gas envisioned by physicists and understudies since it would be a lot simpler if things like intermolecular powers don't exist to confound the basic Ideal Gas Law. Ideal gases are basically point masses moving in consistent, irregular, straight-line movement. Its conduct is portrayed by the suppositions recorded in the Kinetic-Molecular theory of gases.

Complete answer:
We have to know the values of density and volume, to calculate the mass.
The density estimation is utilized to indicate and depict an unadulterated substance, yet in addition decides convergences of parallel combinations and consequently gives data about the piece of blends.
Density = 0.8gm/mL0.8gm/mL ,
Volume = 0.28×103mL0.28 \times {10^3}mL
The formula for calculating mass has to be given below,
d=mVd = \dfrac{m}{V}
Now we can substitute the known values we get,
0.8gm/mL = m0.28×103mL0.8gm/mL{\text{ = }}\dfrac{m}{{0.28 \times {{10}^3}mL}}
m=(0.8gm/mL)×(0.28×103mL)m = \left( {0.8gm/mL} \right) \times \left( {0.28 \times {{10}^3}mL} \right)
Therefore,
m=224gmm = 224gm
Then, we have to calculate the number of moles by using the mass.
n=MassMolarmass=224g28g/mol=8moln = \dfrac{{Mass}}{{Molarmass}} = \dfrac{{224g}}{{28g/mol}} = 8mol
Therefore, the number of moles is 8mol8mol .
By using the ideal gas law,
PV=nRTPV = nRT
Where,
Pressure = 5 atm5{\text{ atm}} ,
Temperature = (27C+273K)=300K\left( {27^\circ C + 273K} \right) = 300K
Universal gas constant = R=0.08atm.litre/K.moleR = 0.08{\text{atm}}{\text{.litre/K}}{\text{.mole}} ,
Applying given values in the above equation to calculate volume.
(5×V)=(8 × 0.08 × 300)\left( {5 \times V} \right) = \left( {8{\text{ }} \times {\text{ 0}}{\text{.08 }} \times {\text{ 300}}} \right)
Therefore,the volume is 38.4mL38.4mL .

Hence, option (B) is correct.

Note:
We have to know that an ideal gas comprises countless indistinguishable atoms. As we know that the particles submit to Newton's laws of movement, and they move in arbitrary movement. We need to know that the atoms experience powers just during impacts; any crashes are totally flexible, and take an unimportant measure of time.