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

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

#### 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

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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