Question

One mole of an ideal gas whose pressure changes with volume as P = al, where a is a
constant, is expanded so that its volume increases n times. Find work done and change in
internal energy. [Adiabatic exponent = y)​

Answer 1

Hence the change in internal energy is dU = αV^2 / γ − 1(η^2 − 1)

Explanation:

Let V be the initial volume of the gas. It expanded to a volume μV . The work done in this process is given by

W=∫ηv - v PdV = ∫ηv - v αVdV = α[V2 / 2]ηv - v  αV2 / 2 η^2 − 1]

The pressure of the gas varies volume as P=αV So, the initial and final pressure will be αV and ηαV .

The change in internal energy is given by

dU = nCVdT = R(Tf − T) / γ − 1 = PfVf −  PiVi / γ −1

dU = η^2 αV^2 −  αV^2 / γ−1 = αV^2 / γ − 1(η^2 − 1)

The heat exchange in this process is given by

Q = U + W

αV^2 / γ−1 [η^2 − 1] + αV^2 / 2[η^2 − 1]

Q  = αV^2 / 2[η^2 − 1] [γ+1 / γ−1]

Here Ti = PiVi / nR = αV^2 / n R and Tf = PfVf / nR = μ^2αV^2 / nR

Now heat capacity C = Q / Tf − T1

C = 1 / Tf − T1 [αV^2 / 2 (η^2 − 1) {γ+1 / γ−1}]

C =  nR / αV^2 (μ^2 − 1)  [αV^2 / 2 (η^2 − 1) {γ+1 / γ−1}]

C = nR^2 [γ+1 / γ−1]

n = 1

Hence the change in internal energy is dU = αV^2 / γ − 1(η^2 − 1)

Answer 2

Explanation:

We know that work done in an adiabatic process is

$=\frac{P_1V_1-P_2V_2}{\gamma-1}$

Given:

Let initial volume be V.

$P_1=aV\\V_2=nV\\\therefore P_2=naV\\\text{Work}=\frac{an^2V^2-anV^2}{\gamma-1} \\=anV^2(\frac{n-1}{\gamma-1})$

Now, change in  internal energy of an adiabatic process

$\Delta U =-\Delta W [\because \Delta Q=0]\\\Delta U =-anV^2(\frac{n-1}{\gamma-1} )$