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The Ratio of the Molar Heat Capacities of an Ideal Gas is Cp/Cv = 7/6. - Physics

Answer in Brief

The ratio of the molar heat capacities of an ideal gas is Cp/Cv = 7/6. Calculate the change in internal energy of 1.0 mole of the gas when its temperature is raised by 50 K (a) keeping the pressure constant (b) keeping the volume constant and (c) adiaba

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Solution

Given:

`(("C""p")/("C""v")) = 7/6`

Number of moles of the gas, n = 1 mol
Change in temperature of the gas, ∆T = 50 K   
(a) Keeping the pressure constant:
Using the first law of thermodynamics,

dQ= dU+dW

`triangle"T" = 50 "K" and gamma  = 7/6`

dQ =dU +dW

work done , dW =PdV

As pressure is kept constant, work done = P(Δ V)

Using the ideal gas equation PV =nRT,

P(ΔV) =nR(ΔT)

⇒ dW =nR(ΔT)

At constant pressure, dQ=nCpdT

Substituting these values in the first law of thermodynamics, we get

nCp dT = dU + RdT
⇒ dU = nCp dT − RdT

Using `"C"^"p"/("C""v") = gamma and "C"_"p" -"C"_"v" = "R", we get`

`"d""U" = 1 xx ("R"gamma)/((gamma -1)) xx "d""T" -"R""d""T"`

 = 7 RdT − RdT
   = 7 RdT − RdT = 6 RdT
   = 6 × 8.3 × 50 = 2490 J

(b) Keeping volume constant:
Work done = 0
Using the first law of thermodynamics,
dU  = dQ
dU = nCv dT

`= 1 xx "R"/(gamma-1) xx "d""T"`

`= 1 xx (8.3/(7/6 - 1)) xx 50`

= 8.3 × 50 × 6 = 2490 J

(c) Adiabatically, dQ = 0
Using the first law of thermodynamics, we get
dU = − dW

=`[("n" xx "R")/(gamma - 1) ("T"_1-"T"_2)]`

= `(1 xx 8.3)/(7/6 -1) = ("T"_2 -"T"_1)`

= 8.3 × 6 ×  50 =2490 J

Concept: Kinetic Theory of Gases and Radiation - Introduction of Kinetic Theory of an Ideal Gas
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APPEARS IN

HC Verma Class 11, Class 12 Concepts of Physics Vol. 2
Chapter 5 Specific Heat Capacities of Gases
Q 5 | Page 77
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