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

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.