Question

Number of \[{{\mathbf{N}}_{\mathbf{2}}}\] molecule present in one-liter vessel at NTP when compressibility factor is \[{\mathbf{1}}.{\mathbf{2}}\] is-
A. \[{\mathbf{2}}.{\mathbf{23}}{\text{ }} \times {\mathbf{1}}{{\mathbf{0}}^{{\mathbf{24}}}}\]
B. \[{\mathbf{2}}.{\mathbf{23}}{\text{ }} \times {\mathbf{1}}{{\mathbf{0}}^{{\mathbf{22}}}}\]
C. \[\;{\mathbf{2}}.{\mathbf{7}}{\text{ }} \times {\mathbf{1}}{{\mathbf{0}}^{{\mathbf{22}}}}\]
D. \[{\mathbf{2}}.{\mathbf{7}}{\text{ }} \times {\text{
}}{\mathbf{1}}{{\mathbf{0}}^{{\mathbf{24}}}}\]

Answer 1

Hint: in the given question compressibility factor \[1.2\] is given of nitrogen gas molecule at normal temperature and pressure which is present in one-liter vessel. We can calculate the number of nitrogen gas molecules by using the formula of compression factor.

Complete Step by step answer: The compressibility factor is denoted by \[Z\], and is also known as the compression factor or can say as the gas deviation factor, which is stated as the ratio of the molar volume of a gas to the molar volume of an ideal gas occur at the same temperature and pressure.
Therefore, the Formula is \[z = \dfrac {{PV}}{{\;nRT}}\]
The normal temperature and pressure shortened as NTP is usually used as a standard condition for testing and citations of fan capacities that is NTP - Normal Temperature and Pressure - is defined as air at $20^\circ $ C and \[1\] atm that is \[101.325{\text{ }}kPa\] and Density \[1.204{\text{ }}kg/{m^3}\;\] that will be \[0.075\;\] pounds per cubic foot
Therefore, form rearranging the formula we get;
\[PV{\text{ }} = {\text{ }}ZnRT\]
We have the values
pressure \[P = {\text{ }}1{\text{ }}atm\]
Volume \[V = 1L\]
compressibility factor \[Z = {\text{ }}1.2\]
universal gas constant \[R = {\text{ }}0.082\]
Temperature \[T = 273K\]
By putting the values in the formula we get;
\[n = \dfrac {{PV}}{{zRT}}\;\]\[ = {\text{ }}\dfrac {{101.325{\text{ }} \times {\text{ }}1 \times 0.001}}{{1.2{\text{ }} \times {\text{ }}8.314{\text{ }} \times {\text{ }}293.15}}\;\]
\[n = {\text{ }}0.0346\]
We know that
\[1\]mole of nitrogen contains \[{N_{2}}{\text{ }} = {\text{ }}6.023 \times {10^{23}}\;\]molecules
From \[PV{\text{ }} = {\text{ }}ZnRT\]
So, \[0.0346\;\]mole contains = \[0.037{\text{ }} \times {\text{ }}6.023{\text{ }} \times {\text{ }}{10^{23}}\;\]
No. of molecules = $0.223 \times {10^{23}}$
No. of molecules = $2.23 \times {10^{22}}$ molecules.

Hence, option B is correct.

Note: The compressibility factor is beneficial for the thermodynamic formula used for changing the ideal gas law to account for behavior of real gas. the ideal gas always has a value of\[1\], and for real gases, the value may deviate positively or negatively.