Thermodynamics Concept Page - 10

Definition

V-T relation for an adiabatic process

For an adiabatic process, PVγ=constant$P{V}^{\gamma }=constant$. Using the ideal gas equation and substituting V=nRTP$V={\displaystyle \frac{nRT}{P}}$ in the above equation gives TVγ−1=constant$T{V}^{\gamma -1}=constant$ where γ$\gamma$ is the specific heat ratio.

Example :
Certain perfect gas is found to obey PV3/2=constant$P{V}^{3/2}=constant$ during an adiabatic process. If such a gas at initial temperature T$T$ is adiabatically compressed to half the initial volume, its final temperature is found as follows:Using the ideal gas law, P$P$ is directly proportional to TV${\displaystyle \frac{T}{V}}$
Substituting this value,
TV0.5=constant$T{V}^{0.5}=constant$
Thus, when the volume is halved, we raise the temperature by 2√$\sqrt{2}$ times.

Definition

P-T relation for an adiabatic process

For an adiabatic process, PVγ=constant$P{V}^{\gamma }=constant$. Using the ideal gas equation and substituting V=nRTP$V={\displaystyle \frac{nRT}{P}}$ in the above equation gives P1−γTγ=constant$P^{1-\gamma }{T}^{\gamma }=constant$ where γ$\gamma$ is the specific heat ratio.

Example:
During an adiabatic process, the pressure of a gas is proportional to the cube of its adiabatic temperature. The value of  Cp/Cv$C_p/C_v$ for that gas is found as follows:Given PαT3→(1)$P\alpha T^3\to (1)$
in adiabatic process relation between P and T is :
P1−γTγ=constant$P^{1-\gamma }{T}^{\gamma }=constant$
Or PαTγγ−1→(2)$P\alpha T^{{\displaystyle \frac{\gamma }{\gamma -1}}}\to (2)$
And we know, for a gas CpCv=γ${\displaystyle \frac{C_p}{C_v}}=\gamma$
From equation 1 and 2:
γγ−1=3${\displaystyle \frac{\gamma }{\gamma -1}}=3$
So, γ=32$\gamma ={\displaystyle \frac{3}{2}}$
So, CpCv=32${\displaystyle \frac{C_p}{C_v}}={\displaystyle \frac{3}{2}}$

Diagram

PV plot for an adiabatic process

Example

VT Plot for adiabatic Process and its variation variation with ratio of specific heat capacities

Diagram

PT plot of adiabatic Process and its variation with ratio of specific heat capacities

Example

Work done in an adiabatic process

Work done in an adiabatic process is given as:
Wb=P1Vγ1V1−γ2−Vγ11−γ=−fnRT1((V2V1)1−γ−1)=−fnRT1((P2P1)γ−1γ−1)$W_b=P_1{V}_1^{\gamma }{\displaystyle \frac{V_2^{1-\gamma }-V_1^{\gamma }}{1-\gamma }}=-fnR{T}_1({({\displaystyle \frac{V_2}{V_1}})}^{1-\gamma }-1)=-fnR{T}_1({({\displaystyle \frac{P_2}{P_1}})}^{{\displaystyle \frac{\gamma -1}{\gamma }}}-1)$
where f is the degree of freedom, γ=f+1f$\gamma ={\displaystyle \frac{f+1}{f}}$ and the subscripts 1 and 2 represent initial and final states respectively.

Example :
5.6$5.6$ liter of helium gas at STP is adiabatically compressed to 0.7$0.7$ liter. Taking the initial temperature to be T1$T_1$, the work done in the process is found as follows:TVγ−1=C$T{V}^{\gamma -1}=C$
so for both volume
T1(5.6)23=T2(0.7)23$T_1(5.6{)}^{{\displaystyle \frac{2}{3}}}=T_2(0.7{)}^{{\displaystyle \frac{2}{3}}}$
or, T2=4T1$T_2=4T_1$
Work done on system =nRΔTγ−1=92RT1${\displaystyle ={\displaystyle \frac{nR\mathrm{\Delta }T}{\gamma -1}}={\displaystyle \frac{9}{2}}R{T}_1}$

Example

Adiabatic process

The relation between U,P$U,P$ and V$V$ for an ideal gas in an adiabatic process is given by relation U=a+bPV$U=a+bPV$. Find the value of adiabatic exponent γ$\gamma$ of this gas.
Solution: In an adiabatic process, according to the first law of thermodynamics
dQ=0=dU+PdV$dQ=0=dU+PdV$
∴dU=−PdV$\therefore dU=-PdV$  ....(1)
Given:
U−a−bPV=0$U-a-bPV=0$
∴dU=bPdV+bVdP$\therefore dU=bPdV+bVdP$  ...(2)
Comparing (1) and (2) 
−PdV=bPdV+bVdP$-PdV=bPdV+bVdP$
∴(b+1)PdV=−bVdP$\therefore (b+1)PdV=-bVdP$
∴dPdV=−(b+1b)PV$\therefore {\displaystyle \frac{dP}{dV}}=-({\displaystyle \frac{b+1}{b}}){\displaystyle \frac{P}{V}}$    ...(3)
Comparing (3) with the equation of adiabatic process
γ=−VPdPdV$\gamma =-{\displaystyle \frac{V}{P}}{\displaystyle \frac{dP}{dV}}$
we get
γ=b+1b$\gamma ={\displaystyle \frac{b+1}{b}}$

Example

Adiabatic mixing of gases

Two closed identical conducting containers are found in the laboratory of an old scientist. For the verification of the gas some experiments are performed on the two boxes and the results are noted 
Experiment 1: When the two containers are weight 
WA$W_A$ =225 g$225g$
WB$W_B$ = 160 g$160g$ and mass of evacuated container
WC$W_C$ = 100 g$100g$.
Experiment 2:
When two containers are given same amount of heat same temperature rise is recorded. The pressure changes are found:
△PA$\mathrm{△}{P}_A$ =2.5 atm$=2.5atm$, △P$\mathrm{△}{P}_{}$ =1.5 atm$=1.5atm$

Required data for unknown gas: Mono (molar gas)
He=4g,Ne=20g,Ar=40g,Kr=84g,Xe=131g,Rd=222g$He=4g,Ne=20g,Ar=40g,Kr=84g,Xe=131g,Rd=222g$

Diatomic (molar gas)
H2=2g,F2=19g,N2=28g,O2=32,Cl2=71g$H_2=2g,F_2=19g,N_2=28g,O_2=32,C{l}_2=71g$

If the gases have initial temperature 300 K$300K$ and they are mixed in an adiabatic container having the same volume as the previous containers. Now the temperature of the mixture is T$T$ and pressure is P$P$. Then

Let the temperature after mixing be Tf$T_f$

nACVA300+nBCVB300=nACVATf+nBCVBTf$n_A{C}_{VA}300+n_B{C}_{VB}300=n_A{C}_{VA}{T}_f+n_B{C}_{VB}{T}_f$⇒Tf=300K$\Rightarrow T_f=300K$ remains the same.
Volume V$V$ remains constant.As, PBV=nBRT$P_{B}V=n_{B}RT$PfV=(nA+nB)RT$P_{f}V=(n_A+n_B)RT$⇒Pf/PB=(nB+nA)/nB>1$\Rightarrow P_f/P_B=(n_B+n_A)/n_B>1$⇒Pf increases$\Rightarrow P_{f}increases$

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