
In an experiment during the analysis of a carbon compound,$145\,L$of ${{H}_{2}}$was collected at $760\,mm\,\,Hg$pressure and ${{27}^{o}}C$temperature. The mass of ${{H}_{2}}$is nearly [MNR$1987$]
A.$10\,\,g$
B.$12\,\,g$
C.$24\,\,g$
D.$6\,\,g$
Answer
164.7k+ views
Hint: According to the ideal law of gas, at high temperature and pressure, the product of the pressure and volume of one gram molecule of an ideal gas equals the product of the temperature of the gas and the universal gas constant. Here the mass ${{H}_{2}}$can be calculated by using the ideal gas law.
Formula Used:The mathematical expression of $n$a mole of ideal gases is:
i)$PV=nRT$
Here $P=$Pressure of the gas
$V=$Volume
$R=$Universal gas constant
$T=$Absolute temperature
ii)$n=\dfrac{m}{M}$
Here $m=$mass of gas
$M=$The molecular mass of the gas
Complete step by step solution:The ideal gas equation gives the relation between the pressure$(P)$, Volume$(V)$, and absolute temperature$(T)$in the limit of high temperature and low pressure such that the molecules of the gas move almost independently of each other. In such a case all gasses obey the ideal gas law or perfect gas law:$PV=nRT$ where $n$denotes the number of moles of ideal gasses.
Here pressure,$P=1\,\,atm$, Volume,$V=145\,\,L$ , Temperature,$T={{27}^{o}}C=(273+27)=300K$
As we know universal gas constant,$R=0.082\,L\,atm\,/mol/K$
Putting these values in an ideal gas equation,
$1\,\,atm\times 145\,\,L\,=n\times 0.082\,\,L\,\,atm/mol/K\times 300K$
Or,$n=\dfrac{1\,\,atm\times 145\,\,L}{0.082\,L\,\,atm/mol/K\,\times 300\,K}=5.8\,mol$
Molar mass of ${{H}_{2}}\,(M)=2g/mol$
$\therefore $Mass of ${{H}_{2}}(m)=n\times M$
Or,$m=5.8mol\times 2g/mol=11.6g\approx 12g$
Therefore the mass of ${{H}_{2}}$is nearly equal to $12g$.
Thus, option (B) is correct.
Note: Ideal gas behaves as real gas at high pressure and low temperature. According to ideal gas law, there is no or negligible molecular interaction between gas molecules but real gasses do have molecular interaction under specific conditions. If we increase the density of ideal gas very much such that gasses become very close to close and start to interact with each other.
Formula Used:The mathematical expression of $n$a mole of ideal gases is:
i)$PV=nRT$
Here $P=$Pressure of the gas
$V=$Volume
$R=$Universal gas constant
$T=$Absolute temperature
ii)$n=\dfrac{m}{M}$
Here $m=$mass of gas
$M=$The molecular mass of the gas
Complete step by step solution:The ideal gas equation gives the relation between the pressure$(P)$, Volume$(V)$, and absolute temperature$(T)$in the limit of high temperature and low pressure such that the molecules of the gas move almost independently of each other. In such a case all gasses obey the ideal gas law or perfect gas law:$PV=nRT$ where $n$denotes the number of moles of ideal gasses.
Here pressure,$P=1\,\,atm$, Volume,$V=145\,\,L$ , Temperature,$T={{27}^{o}}C=(273+27)=300K$
As we know universal gas constant,$R=0.082\,L\,atm\,/mol/K$
Putting these values in an ideal gas equation,
$1\,\,atm\times 145\,\,L\,=n\times 0.082\,\,L\,\,atm/mol/K\times 300K$
Or,$n=\dfrac{1\,\,atm\times 145\,\,L}{0.082\,L\,\,atm/mol/K\,\times 300\,K}=5.8\,mol$
Molar mass of ${{H}_{2}}\,(M)=2g/mol$
$\therefore $Mass of ${{H}_{2}}(m)=n\times M$
Or,$m=5.8mol\times 2g/mol=11.6g\approx 12g$
Therefore the mass of ${{H}_{2}}$is nearly equal to $12g$.
Thus, option (B) is correct.
Note: Ideal gas behaves as real gas at high pressure and low temperature. According to ideal gas law, there is no or negligible molecular interaction between gas molecules but real gasses do have molecular interaction under specific conditions. If we increase the density of ideal gas very much such that gasses become very close to close and start to interact with each other.
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