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Question: The density of gas is found to be \[2.07g{L^{ - 1}}\] at \[{30^ \circ }C\] and \[2\] atmospheric pre...

The density of gas is found to be 2.07gL12.07g{L^{ - 1}} at 30C{30^ \circ }C and 22 atmospheric pressure. What is its density at NTP?

Explanation

Solution

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 293.15K293.15Kand 11 atmospheric pressure.

Complete answer:
In the question it is given:
Density (D1{D_1}) -2.07gL12.07g{L^{ - 1}}, temperature (T1{T_1})-303K303K, pressure(P2{{\text{P}}_2})-22atm
As the molecular mass of gas remains constant applying the formula, of ideal gas law:
PV=nRT\Rightarrow PV = nRT
P=nVRT\Rightarrow P = \dfrac{n}{V}RT
P=mMVRT\Rightarrow P = \dfrac{m}{{MV}}RT
P1=D1MRT1\Rightarrow {P_1} = \dfrac{{{D_1}}}{M}R{T_1} and P2=D2MRT2{{\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: T2{T_2}-298K298K, P2{{\text{P}}_2}-1atm1atm
P1D1T1=P2D2T2\Rightarrow \dfrac{{{P_1}}}{{{D_1}{T_1}}} = \dfrac{{{P_2}}}{{{D_2}{T_2}}}
Substituting the values of given parameters to find D2{D_2}
D2=1×2.07×3032×298{D_2} = \dfrac{{1 \times 2.07 \times 303}}{{2 \times 298}}
Solving we have
D2=1.49gL1\Rightarrow {D_2} = 1.49g{L^{ - 1}}
Thus, the density of gas at NTP is 1.49gL11.49g{L^{ - 1}}.

Note:
One more standard condition defined by IUPAC is the STP (Standard Temperature and Pressure) condition in which the temperature is 273.15K273.15K and the pressure condition is 0.9870.987atmospheric pressure. The condition for STP and NTP varies in difference in temperature by 20K20Kand 0.013atm0.013atm