Department of Pre-University Education, KarnatakaPUC Karnataka Science Class 11

Half Mole of an Ideal Gas (γ = 5/3) is Taken Through the Cycle Abcda, as Shown in the Figure. Take R = 25 3 J K − 1 Mol − 1 . - Physics

Half mole of an ideal gas (γ = 5/3) is taken through the cycle abcda, as shown in the figure. Take  "R" = 25/3"J""K"^-1 "mol"^-1 . (a) Find the temperature of the gas in the states a, b, c and d. (b) Find the amount of heat supplied in the processes ab and bc. (c) Find the amount of heat liberated in the processes cd and da.

Solution

Given:
Number of moles of the gas,

n = 0.5 mol

"R" = 25/3 J/mol -K

gamma =5/3

(a) Temperature at a = Ta

PaVa =nRTa

=> "T""a" = ("P"_"a""V"_"a")/("n""R") = (100 xx 10^3 xx 5000 xx 10^-6)/(0.5 xx 25/3) = 120 "K"

Similarly, temperature at b,

"T"_"b" = ("P"_b"V"_"b")/("n""R")

"T"_"b" =( 100 xx 10^3 xx 10000 xx 10^-6)/(0.5 xx 25/3)

Tb =240 K

Similarly, temperature at c  is 480 K and at d  is 240 K.
(b) For process ab,
dQ = ncpdT
[Since ab is isobaric]

"d""Q" = 1/2 xx ("R"gamma)/(gamma-1) ("T"_"b" - "T"_"a")

"d""Q" =1/2 xx ((25 xx 5)/(3 xx 3))/(5/3 -1 )  xx (240 -120 )

"d""Q"  = 1/2 xx 125 /9 xx 3/2 xx (120)

dQ = 1250 J

For line bc, volume is constant. So, it is an isochoric process.
dQ = dU + dW
[dW = 0, isochoric process]
dQ = dU = nCvdT

dQ = nCv (Tc - Tb)

"d""Q" = 1/2 xx ((25/3))/[[(5/3)-1]] xx (240)

"d""Q" = 1/2 xx 25/3  xx 3/2 xx 240 =1500 "J"

(c) Heat liberated in cd (isobaric process),
dQ = − nCpdT

"d""Q" = -1/2 xx( gamma "R")/(gamma-1) xx ("T"_d -"T"_"c")

"d""Q" = -1/2  xx 125/9 xx 3/2 xx ( 240 -480)

"d""Q" = -1/2 xx 125/6  xx 240 = 2500 "J"

Heat liberated in da (isochoric process),
⇒ dQ = dU
Q= −nCvdT

"d""Q" = -1/2 xx  "R" /(gamma -1 )("T"_"a"-"T"_"d")

"d""Q" = -1/2 xx 25/2 xx (120-240)

"d""Q" = 25/4 xx 120 =750 "J"

Concept: Kinetic Theory of Gases and Radiation - Introduction of Kinetic Theory of an Ideal Gas
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HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 5 Specific Heat Capacities of Gases
Q 13 | Page 78
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