KINETIC THEORY OF GASES 2-2: What is the mean kinetic energy of individual nitro
ID: 1409255 • Letter: K
Question
KINETIC THEORY OF GASES 2-2:
What is the mean kinetic energy of individual nitrogen molecules at 1500 K in Joules? In electron volts?
KINETIC THEORY OF GASES 2-8:
One mole of oxygen is heated at a constant pressure starting at 0 C. How much heat energy must be added to the gas to double its volume?
ENTROPY AND THE SECOND LAW OF THERMODYNAMICS 2-22:
In a Carnot cycle, the isothermal expansion of an ideal gas takes place at 400 K and the isothermal compression at 300 K. During the expansion 500 cal of heat energy are transferred to the gas. Determine (a) the work performed by the gas during the isothermal expansion, (b) the heat rejected from the gas during the isothermal compression, (c) the work done on the gas during the isothermal compression.
ENTROPY AND THE SECOND LAW OF THERMODYNAMICS 2-30:
In a specific heat experiment, 200 g of aluminium (Cp = 0.215 cal/g.°C) at 1000°C is mixed with 50 g of water at 20°C. Find the difference in entropy of the system at the end from its value before mixing.
Explanation / Answer
1)
KE = (3/2) kb T
= (3/2) . 1.38064810?23 . 1500
= 3.105*10-20 J (a)
joule = 6.2415 1018 electron volts
Therefore, 2.8993510-20 J = 3.105*10-20. 6.2415 1018 eV = 0.193752 eV
2)
Q = n Cp DT
n = 1
T1 = 0 °C = 273 K
T2 = ? K
V1 = V
V2 = 2 V
p1 = p2 = p
p V = n R T
T2 / T1 = V2 / V1 = 2
T = 2 T1
DT = T2 - T1 = 2 T1 - T1 = T1 = 273 K
Cp = CV + R
CV = ( f / 2 ) R
degrees of freedom (translation + rotation), f = 3 + 2 = 5
Cp = ( 5/2 + 1) R = ( 7 / 2 ) R
R = 8.31 J.mol-1.K-1
Q = n Cp DT = (1)(7/2)(8.31)(273)
Q = 7.9x 10^3 J
3)
nR(T1-T2)ln(Va/Vb) is work done which is equal to : 500*100/400 = 125J
Work performed by gas = 375J
Efficiency = 125/500 = 0.25
4)
We know 1 Cal = 4.18 Joules
200*0.215*4.18*(1273-T) + 50*4.18(T-293)
T = 431K
entropy difference = Q/T = 351.11 J
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