Answer 31)
(a) Given: P1= 1 atm = 1.013x105 Pa, T1 = 20°C =T, V1 =1.0 cm3 = V2 =V.
Find: (a) N1 = n1/NA , NA = 6.02 molecules/mole. (b) Given: P2 = 1.0 x 10-11 Pa , T2 = T1 = 20°C =T = T,
V2 =V1 = V= 1.0 cm3 .
Find: n2
Conversions V2 =V1 = V= 1.0 cm3 1.0 [cm]3 = 1.0 [cm( )m/cm]3 = 1.0 [cm(10-2)m/cm]3 = 1.0 [(10-2)m]3 = 1.0 x10-6 m .
T2 = T1 = T = 20°C = 273.15 K + 20 = 293.15 K.
Physical Principle Ideal Gas Law: PV = nRT ------------eq(1)
R = the Universal Gas Constant = 8.31 J/mole K.
(a)
T1 = T = 293.15 K T and V1 = V = 1.0 x10-6 m
n1 = P1V1/(RT1) = P1V1/(RT) ---------------eq(2)
= 1.013x10-5 Pa *1.0 x10-6 m3 /(8.31 J/mole K*293.15 K) = 4.2x10-5 moles
N1 = n1/NA = 4.2x10-5 moles/6.02 molecules/mole = 2.5x1019 molecules.
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(b) Use Eq. (2) now for n2 with T2 = T = 293.15 K T and V2 = V = 1.0 x10-6 m gives
n2 = P2V2/(RT2 ) = P2V/(RT)-------------------------eq (3)
= 1.0 x 10-11 Pa x 1.0 x10-6 m3 /(8.31 J/mole K*293.15 K)
=4.1 x10-21 mol.
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Answer 33)
let there be (n) gram mole of gas in tank initially
P1V = n R T1
75° C=348K
25° C=298K
----------------------
2n/3 moles is released, so left in tank in (n/3) moles, V = tank
same
P2V = (n/3) R T2
-----------------------------
divide
[P2/P1] = [1/3][T2/T1]
= (1/3)[348 K/298K] = 0.3893
P2 = 0.3893*P1 = 11*0.3893 = 4.28 atm
31 and 33 Draw a diagram for each of processes (isothermal, isobaric, isochoric) in variables (P,...
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