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Question
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
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