The ratio of the molar heat capacities of an ideal gas is C_{p}/C_{v} = 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=nC_{p}dT

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

nC_{p}_{ }dT = dU + RdT

⇒ dU = nC_{p} 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 = nC_{v} 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