Question
Question: One half moles each of nitrogen, oxygen and carbon dioxide are mixed in an enclosure of volume 5 lit...
One half moles each of nitrogen, oxygen and carbon dioxide are mixed in an enclosure of volume 5 liters and temperature 27oC. The pressure exerted by the mixture is (R=8.31J/mol/K)
(A) 7.48×105Nm−2
(B) 4×105Nm−2
(C) 6×105Nm−2
(D) 3×105Nm−2
Solution
Apply Dalton’s law of partial pressure which gives the pressure of the mixture altogether. Pressure is equal to the ratio of temperature, number of moles, gas constant to the volume, P=VnRT . Substitute the data and calculate the pressure of the mixture.
Complete step-by-step solution
According to Dalton’s law of partial pressure, the total pressure exerted by a mixture of non reacting gases occupying a vessel is equal to the sum of individual pressures which each gases exert if it alone occupied the same volume at given temperature.
For n gases, P=P1+P2+P3+...Pn
Let P1,P2,P3 be pressures exerted by nitrogen, oxygen and carbon dioxide respectively.
From ideal gas equation,
$
PV = nRT \\
P = {P_{{N_2}}} + {P_{{O_2}}} + {P_{C{O_2}}} \\
P = \dfrac{{nRT}}{V} + \dfrac{{nRT}}{V} + \dfrac{{nRT}}{V} \\
P = \dfrac{{3nRT}}{V} \\
Given, the number of moles of any particular gas $${\text{n = 0}}{\text{.5}}$$ moles; temperature $${\text{T = 2}}{{\text{7}}^{\text{o}}}{\text{C = 300K}}$$; Volume:V = 5 \times 10^{-3} m^3$
Substitute the data in the expression-
$
P = \dfrac{{3 \times 8.31 \times 300}}{{2 \times 5 \times {{10}^{ - 3}}}} \\
P = 7.48 \times {10^5}N{m^{ - 2}} \\
Hence,thepressureexertedbythemixtureis7.48 \times {10^5}N{m^{ - 2}}$.
The correct option is A.
Note: The relation given below is a way to determine a volume based concentration of the individual gases.
P=Ptotalc
c is the concentration of the component.
Real gases do not obey Dalton’s law with deviation increasing with the pressure. Then volume occupied by the molecules becomes more significant than free space between them.