Question

The density of gas is found to be \[2.07g{L^{ - 1}}\] at \[{30^ \circ }C\] and \[2\] atmospheric pressure. What is its density at NTP?

Answer 1

Hint: IUPAC has defined some of the conditions as the standard conditions to perform any experimental measurements so that various experiments can be compared with each other. NTP condition for gas is normal temperature and pressure in which temperature is \[293.15K\]and \[1\] atmospheric pressure.

Complete answer:
In the question it is given:
Density (\[{D_1}\]) -\[2.07g{L^{ - 1}}\], temperature (\[{T_1}\])-\[303K\], pressure(\[{{\text{P}}_2}\])-\[2\]atm
As the molecular mass of gas remains constant applying the formula, of ideal gas law:
\[ \Rightarrow PV = nRT\]
\[ \Rightarrow P = \dfrac{n}{V}RT\]
\[ \Rightarrow P = \dfrac{m}{{MV}}RT\]
\[ \Rightarrow {P_1} = \dfrac{{{D_1}}}{M}R{T_1}\] and \[{{\text{P}}_2} = \dfrac{{{D_2}}}{M}R{T_2}\]
Where: P is pressure, V-volume , n-no of moles, R is gas constant, T-temperature, m- mass of gas, D is density, M is molecular mass of gas.
At NTP: \[{T_2}\]-\[298K\], \[{{\text{P}}_2}\]-\[1atm\]
\[ \Rightarrow \dfrac{{{P_1}}}{{{D_1}{T_1}}} = \dfrac{{{P_2}}}{{{D_2}{T_2}}}\]
Substituting the values of given parameters to find \[{D_2}\]
\[{D_2} = \dfrac{{1 \times 2.07 \times 303}}{{2 \times 298}}\]
Solving we have
\[ \Rightarrow {D_2} = 1.49g{L^{ - 1}}\]
Thus, the density of gas at NTP is \[1.49g{L^{ - 1}}\].

Note:
One more standard condition defined by IUPAC is the STP (Standard Temperature and Pressure) condition in which the temperature is \[273.15K\] and the pressure condition is \[0.987\]atmospheric pressure. The condition for STP and NTP varies in difference in temperature by \[20K\]and \[0.013atm\]