1) A closed steel container with an inner volume of 5.10 m 3contains s.oo m 3 of
ID: 1872061 • Letter: 1
Question
1) A closed steel container with an inner volume of 5.10 m 3contains s.oo m 3 of gasoline and o.10 m 3 of nitrogen all at 2ooc and 1.00 atm of pressure. (a) If the temperature of the container and contents are raised to 6o °C what will be the volume of the container? (b) What will be the volume of the gasoline? (C) What will be the pressure of the nitrogen? steel linear thermal expansion coefficient 12.2 x 10-6 K gasoline volume thermal expansion coefficient 950x 106K (Neglect the vapor pressure of the gasoline.)Explanation / Answer
Given
closed steel container with
volume initial v1 = 5.10 m^3
with gasoline vg1 = 5.0 m^3
nitrogen volume is vn1 = 0.1 m^3
initial temperature T1 = 20 0C = 293.15 k
final temperature is T2 = 60 0C = 333.15 k
initial pressure of the system is P1 = 1 atm = 1.01*10^5 Pa
final pressure P2 =?
a) final volume of the container is V2 = ?
as the container made up of steel so upon heating the volume increases that is
V2 = V1(1+beta(dT))
here beta is = 3*alpha , alpha is coefficient of linear expansion
given alpha = 12.2*10^-6 k
v2 = 5.10 (1+3*12.2*10^-6(40)) m^3
v2 = 5.1074664 m^3
the final volume of gasoline is
given alpha = 950*10^-6 k
vg2 = 5.0 (1+3*95*10^-6(40)) m^3
vg2 = 5.057 m^3
so the volume of the nitrogen is Vn2 = V2 - vg2 = 5.1074664 -5.057 = 0.0504664 m^3
that is 0.1+0.0504664 m^3 = 0.1504664 m^3
c) pressure of the nitrogen is PV = nRT
P1V1/T1 = P2V2/T2
P2 = P1V1T2/T1*V2
P2 = 1.01*10^5*0.1*333.15/(293.15*0.1504664) Pa
P2 = 76283.702282110098 Pa
P2 = (76283.702282110098)/(1.01*10^5) atm
P2 = 0.7552842 atm
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