Thermodynamics Concept Page - 11

Example

Adiabatic process in a piston cylinder process

An ideal monoatomic gas is confined in a cylinder by a spring loaded piston of cross section 8.0×10−3m2$8.0\times {10}^{-3}{m}^2$. Initially the gas is at 300 K$300K$ and occupies a volume of 2.4×10−3m3$2.4\times {10}^{-3}{m}^3$ and the spring is in its relaxed state as shown in figure. The gas is heated by a small heater until the piston moves out slowly by 0.1 m$0.1m$. The force constant of the spring is 8000 N/m$8000N/m$ and the atmospheric pressure is 1.0×105N/m2$1.0\times {10}^{5}N/m^2$. The cylinder and the piston are thermally insulated. The piston and the spring are massless and there is no friction between the piston and the cylinder. The final temperature of the gas will be :
(Neglect the heat loss through the lead wires of the heater. The heat capacity of the heater coil is also negligible)F=kx=8000×0.1×800N$F=kx=8000\times 0.1\times 800N$
Pressure exerted on the piston by the spring is:
ΔP=FA=8008×10−3=1×105N/m2$\mathrm{\Delta }P={\displaystyle \frac{F}{A}}={\displaystyle \frac{800}{8\times {10}^{-3}}}=1\times {10}^{5}N/m^2$
The total pressure P2$P_2$ of the gas inside cylinder is              .
P2=Patm+ΔP=1×105+1×105=2×105N/m2$P_2=P_{atm}+\mathrm{\Delta }P=1\times {10}^5+1\times {10}^5=2\times {10}^{5}N/m^2$
Since the piston has moved outwards, there has been an increase of ΔV$\mathrm{\Delta }V$ in the volume of the gas, i.e.,
ΔV=A×x=(8×10−3)×(0.1)$\mathrm{\Delta }V=A\times x=(8\times {10}^{-3})\times (0.1)$
=8×10−4m3$=8\times {10}^{-4}{m}^3$
The final volume of the gas
V2=V1+ΔV=2.4×10−3+8×10−4=3.2×10−3m3$V_2=V_1+\mathrm{\Delta }V=2.4\times {10}^{-3}+8\times {10}^{-4}=3.2\times {10}^{-3}{m}^3$
Let T2$T_2$ be the final temperature of gas. Then
(P1V1/T1)=(P2V2/T2)⇒T2=(P2V2/P1V1)T1$(P_1{V}_1/T_1)=(P_2{V}_2/T_2)\Rightarrow T_2=(P_2{V}_2/P_1{V}_1)T_1$
=300×(2×105×3.2×10−3)/(105×2.4×10−3)$=300\times (2\times {10}^5\times 3.2\times {10}^{-3})/({10}^5\times 2.4\times {10}^{-3})$
=800K$=800K$

Example

Infering PV-plot

Example: Figure shows three processes for an ideal gas. The temperature at a$a$ is 600 K$600K$, pressure 16$16$ atm and volume 1$1$ litre. The volume at b$b$ is 4$4$ litre. Out of the two process ab$ab$ and ac$ac$, one is adiabatic and the other is isothermal. The ratio of specific heats of the gas is 1.5$1.5$. Compute the pressure of the gas at b$b$ and c$c$.

Solution:
The process ab is an adiabatic process.Given :    γ=1.5$\gamma =1.5$ , Pa=16$P_a=16$ atm , Va=1$V_a=1$ Litre           Vb=4$V_b=4$ Litre
Using        PVγ=constant$P{V}^{\gamma }=constant$
⟹$⟹$      PbPa=(VaVb)γ${\displaystyle \frac{P_b}{P_a}}=({\displaystyle \frac{V_a}{V_b}}{)}^{\gamma }$
∴$\therefore$       Pb16=(14)1.5${\displaystyle \frac{P_b}{16}}=({\displaystyle \frac{1}{4}}{)}^{1.5}$OR       Pb16=0.125${\displaystyle \frac{P_b}{16}}=0.125$          
    ⟹Pb=2$⟹P_b=2$ atmAlso    Pc=Pb=2$P_c=P_b=2$ atm

Definition

Work done in a polytropic process

Work done in a polytropic process
Wb=P2V2−P1V11−n$W_b={\displaystyle \frac{P_2{V}_2-P_1{V}_1}{1-n}}$   for n≠1$n\ne 1$
Wb=PVln(V2V1)$W_b=PVln({\displaystyle \frac{V_2}{V_1}})$   for n=1$n=1$

Definition

Polytropic process

A polytropic process is a thermodynamic process that obeys the relation:
pvn=C$p{v}^n=C$
where p$p$ is the pressure, v$v$ is specific volume, n$n$ is the polytropic index (any real number), and C$C$ is a constant.

Definition

PV during a polytropic process

Air (ideal gas with g = 1.4) at 1$1$ bar and 300 K$300K$ is compressed till the final volume is one-sixteenth of the original volume, following a polytropic process PV1.25=$P{V}^{1.25}=$const. In polytropic process PVη=constant$P{V}^{\eta }=constant$
∴P1Vη1=P2Vη2$\therefore P_1{V}_1^{\eta }=P_2{V}_2^{\eta }$    ....(1)
Given,
P1=1$P_1=1$
V2=116V1$V_2={\displaystyle \frac{1}{16}}{V}_1$
η=1.25$\eta =1.25$
Substituting in (1) 1×V1.251=P2×(116)1.25V1.251$1\times V_1^{1.25}=P_2\times ({\displaystyle \frac{1}{16}}{)}^{1.25}{V}_1^{1.25}$
P2=32Pa$P_2=32Pa$

Example

Polytropic process- relation between T and V

Air (ideal gas with g = 1.4) at 1$1$ bar and 300 K$300K$ is compressed till the final volume is one-sixteenth of the original volume, following a polytropic process PV1.25=$P{V}^{1.25}=$const. In polytropic process
TVη−1=constant$T{V}^{\eta -1}=constant$
T1Vη−11=T2Vη−12$T_1{V}_1^{\eta -1}=T_2{V}_2^{\eta -1}$
∴300×V1.251=T2(116)1.25×V1.251$\therefore 300\times V_1^{1.25}=T_2({\displaystyle \frac{1}{16}}{)}^{1.25}\times V_1^{1.25}$
∴T2=300×2=600K$\therefore T_2=300\times 2=600K$

Example

Polytropic process- relation between T and P

A piston-cylinder contains water at 5000C${500}^{0}C$, 3$3$ MPa. It is cooled in a polytropic process to 2000C${200}^{0}C$, 1$1$ MPa.In polytropic process:
TηPη−1=constant${\displaystyle \frac{T^{\eta }}{P^{\eta -1}}}=constant$
∴(500+273200+273)η=(31)η−1$\therefore ({\displaystyle \frac{500+273}{200+273}}{)}^{\eta }=({\displaystyle \frac{3}{1}}{)}^{\eta -1}$
(773473)η=3η−1$({\displaystyle \frac{773}{473}}{)}^{\eta }=3^{\eta -1}$
ηln(773473)=(η−1)ln(3)$\eta \ln ({\displaystyle \frac{773}{473}})=(\eta -1)\ln (3)$
η(0.4911)=(η−1)(1.098)$\eta (0.4911)=(\eta -1)(1.098)$
∴η=1.81$\therefore \eta =1.81$

Diagram

PV plot for a polytropic process

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