Thermodynamics

In adiabatic process
PV? = constant
P (m/ρ)? = constant
As mass is constant
P ∝ ρ?
P_f/P_i = (ρ_f/ρ_i)? = (32)? /? = 2? = 128

  $\mathrm{\Delta }\mathcal{l}=0.03\mathrm{m}\mathrm{m}$ $$

${\mathrm{P}}_1=1\mathrm{a}\mathrm{t}\mathrm{m},{\mathrm{T}}_1=273\mathrm{K}$

Now work done $\begin{array}{rr}& {}_{}{}_{}^{}{}_{}{}_{}^{}\\ \end{array}$ $\begin{array}{rr}& {\mathrm{P}}_1{\mathrm{V}}_1^{\gamma }={\mathrm{P}}_2{\mathrm{V}}_2^{\gamma }\\ & {\mathrm{P}}_2={\mathrm{P}}_1{\left[\frac{{\mathrm{V}}_1}{{\mathrm{V}}_2}\right]}^{\gamma }\\ & =1\mathrm{a}\mathrm{t}\mathrm{m}{\left(\frac{1}{3}\right)}^{1.4}\end{array}$

Closes answer is $\frac{{}_{}{}_{}{}_{}{}_{}}{}$ $=\frac{P_1{V}_1-P_2{V}_2}{\gamma -1}=88.7\mathrm{J}$ .

  $\mathrm{}{\gamma }_{\text{mixture}}=\frac{n_1{C}_{P_1}+n_2{C}_{P_2}}{n_1{C}_{V_1}+n_2{C}_{V_2}}=\frac{n_1\frac{{\gamma }_{1}R}{{\gamma }_1-1}+n_2\frac{{\gamma }_{2}R}{{\gamma }_2-1}}{\frac{n_{1}R}{{\gamma }_1-1}+\frac{n_{2}R}{{\gamma }_2-1}}$ $\mathrm{}{}_{}\frac{{}_{}{}_{{}_{}}{}_{}{}_{{}_{}}}{{}_{}{}_{{}_{}}{}_{}{}_{{}_{}}}\frac{{}_{}\frac{{}_{}}{{}_{}}{}_{}\frac{{}_{}}{{}_{}}}{\frac{{}_{}}{{}_{}}\frac{{}_{}}{{}_{}}}$

on rearranging we get

$\frac{{\mathrm{n}}_1+{\mathrm{n}}_2}{{\gamma }_{\text{mix}}-1}=\frac{{\mathrm{n}}_1}{{\gamma }_1-1}+\frac{{\mathrm{n}}_2}{{\gamma }_2-1};\frac{5}{{\gamma }_{\text{mix}}-1}=\frac{3}{1/3}+\frac{2}{2/3}$

$\frac{5}{{\gamma }_{\text{mix}}-1}=9+3=12\mathrm{}\Rightarrow {\gamma }_{\text{mixture}}=\frac{17}{12}+1+\frac{5}{12};{\gamma }_{\text{mix}}=1.42$

Relaxation time $\left(\tau \right)\propto \frac{\mathrm{V}}{\sqrt[]{\mathrm{T}}}$

and $\mathrm{}\mathrm{T}\propto \frac{1}{{\mathrm{V}}^{\gamma -1}}$

$\begin{array}{rr}& \tau \propto {\mathrm{V}}^{1+\frac{\gamma -1}{2}}\mathrm{}\tau \propto {\mathrm{V}}^{\frac{1+\gamma }{2}}\\ & \frac{{\tau }_f}{{\tau }_{\mathrm{i}}}={\left(\frac{2\mathrm{V}}{\mathrm{V}}\right)}^{\frac{1+\gamma }{2}}\mathrm{}\frac{{\tau }_{\mathrm{f}}}{{\tau }_{\mathrm{i}}}=(2{)}^{\frac{1+\gamma }{2}}\end{array}$

$\mathrm{W}={\mathrm{Q}}_1-{\mathrm{Q}}_2={\mathrm{Q}}_2-{\mathrm{Q}}_3$

${\mathrm{Q}}_2=\frac{{\mathrm{Q}}_1+{\mathrm{Q}}_3}{2}$

$2=\frac{{\mathrm{Q}}_1}{{\mathrm{Q}}_2}+\frac{{\mathrm{Q}}_3}{{\mathrm{Q}}_2}$

$2=\frac{T_1}{T}+\frac{T_2}{T}$

$\mathrm{T}=\frac{{\mathrm{T}}_1+{\mathrm{T}}_2}{2}$

η_A = W_A/Q? = 1 - T/T?
η_B = W_B/Q? ' = 1 - T? /T
W_A = Q? (1 - T/T? ) = Q? - Q?
W_B = Q? ' (1 - T? /T) = Q? /2 (1 - T? /T)
Given W_A = W_B
Q? (1 - T/T? ) = (Q? /2) (1 - T? /T)
Q? (T? /T) (1-T/T? ) = (Q? /2) (1-T? /T)
(T? /T - 1) = (1/2) (1-T? /T)
2T? /T - 2 = 1 - T? /T
2T? /T + T? /T = 3
T = (2T? +T? )/3

Degrees of freedom (f) = 3 translational + 3 rotational + (2 * 4) vibrational = 14
γ = 1 + 2/f = 1 + 2/14 = 8/7
W = nR? T/ (γ-1) = (1 * 8.3 * (310-300)/ (8/7 - 1) = (83)/ (1/7) = 581 J
As W is positive, work is done by the gas. The solution says W<0, work done on the gas. This implies? T is negative. The question states temperature rises, so work is done on the gas. W = -582J.

H? O (l) → H? O (g)
ΔH° = ΔU° + ΔngRT
ΔH° - ΔU° = ΔngRT
= 1 * 8.31 * 373
= 3099.63 J/mol
= 30.9963 * 10² J/mol
≈ 31 * 10² J/mol

-dT/dt = K [T - Ts]

• (61-59)/4 = K [ (61+59)/2 - 30]
  -0.5 = K [60 - 30] = 30K
  So, K = -1/60 min? ¹
  Again
• (51-49)/t = K [ (51+49)/2 - 30]
  -2/t = (-1/60) [50-30] = -20/60 = -1/3
  t = 6 min

A→B & D→C (isothermal process)
So, TA = TB & TD = TC. Now B→C & D→A (adiabatic process)
|WBC| = nR/ (γ-1) (TB - TC)
|WAD| = nR/ (γ-1) (TA - TD) = nR/ (γ-1) (TB - TC)
∴ |WBC| = |WAD|