We have given,

initial pressure of monoatomic gas = 8 atm

initial temperature T = 400k

(a) ΔS= Cp ln T₂/T₁ - Rln P₂/P₁

for isothermal process \(\triangle\)T = 0 i.e T₂ = T₁

\(\therefore\) \(\triangle\)s = -Rln P₂/P₁

 \(\triangle\)s = -8.314 ln 1/8

 \(\triangle\)s = -8.314 x (-2.1)

 \(\triangle\)s = 17.3 Jk^-1

(b) when,  gas decreases its pressure at constant volume to a pressure of 5 atm, there will be also change in temperature.

\(\therefore\) final temperature T₂ = \(\frac{P_2\times T_1}{P_1}\)

 = \(\frac{5atm\times400k}8\) = 250 k

\(\therefore\)  \(\triangle\)s = C_p ln T₂/T₁ - Rln P₂/P₁

 = 5R/2 ln 250/400 - Rln(5/8)

(for monoatomic gas C_p = 5R/2)

 \(\triangle\)s = 5/2 x 8.314 x ln(250/400) - 8.314 ln(5/8)

 = -9.77 jk^-1- (-3.91Jk^-1)

 \(\triangle\)s = -5.86 Jk^-1

(c)  \(\triangle\)u = q + w

for adiabatre process  - q = 0

 \(\triangle\)u = w (w = -ve, expression)

C_v x  (T₂ - T₁) = -400

3/2 R(T₂ - 400k) = -400

(T₂ - 400k) = \(\frac{-400\times2}{3\times8.314}\)

T₂ = 368 k

 \(\triangle\)s = C_pln T₂/T₁ - Rln P₂/P₁

 =  \(\frac52\) 8.314 ln 368/400 - 8.314 ln2/8

 = (-1.73) - (-11.520)

 \(\triangle\)s  = 9.8 Jk^-1