Thermodynamics Concept Page - 5

Example

Problem on first law where cyclic process is defined graphically

Example: Find the heat energy absorbed by a system in going through a cyclic process shown in figure.

Solution:
Since process is cyclic, change in internal energy ΔU=0$\mathrm{\Delta }U=0$
Applying first law of thermodynamics,
ΔQ=ΔU+ΔW$\mathrm{\Delta }Q=\mathrm{\Delta }U+\mathrm{\Delta }W$
ΔQ=ΔW$\mathrm{\Delta }Q=\mathrm{\Delta }W$ = Area under the ellipse= πab$\pi ab$
Here, length of semi major axis=a=10=b=$=a=10=b=$length of semi minor axis
Hence, heat absorbed= 102πJ${10}^2\pi J$

Definition

First Law of thermodynamics for isochoric process

For an isochoric process ΔV=0⇒ΔW=0$\mathrm{\Delta }V=0\Rightarrow \mathrm{\Delta }W=0$
According to the first law; ΔQ=ΔU=ncvΔT$\mathrm{\Delta }Q=\mathrm{\Delta }U=n{c}_v\mathrm{\Delta }T$

Definition

First Law of thermodynamics for isobaric process

For an isobaric process, ΔW=pΔV=nRΔT$\mathrm{\Delta }W=p\mathrm{\Delta }V=nR\mathrm{\Delta }T$ and ΔU=ncvΔT$\mathrm{\Delta }U=n{c}_v\mathrm{\Delta }T$
So according to the first law;
ΔQ=ncvΔT+nRΔT$\mathrm{\Delta }Q=n{c}_v\mathrm{\Delta }T+nR\mathrm{\Delta }T$

Definition

First Law for isothermal process

In an isothermal process ΔT=0⇒ΔU=0$\mathrm{\Delta }T=0\Rightarrow \mathrm{\Delta }U=0$
So according to first law; ΔQ=ΔW$\mathrm{\Delta }Q=\mathrm{\Delta }W$

Example

Solve problems using First Law involving piston and cylinder with constant external pressure

Example: A mass less piston, which can move without friction , closes a sample of Helium in a vertical, thermally insulated cylinder, which is closed at its bottom , and the cross section of which is A=2dm2$A=2d{m}^2$. Above the piston there is a fixed stand to which an unstretched spring of spring constant 2000 N/m$2000N/m$ is attached, whose bottom end is at a distance of 2 dm$2dm$ from the piston when the volume of the gas is V0=8dm3$V_0=8d{m}^3$. The external pressure is p0=105Pa,g=10m/s2$p_0={10}^{5}Pa,g=10m/s^2$. The gas confined in the cylinder is heated with an electric heating element 1dm=10−1m.$1dm={10}^{-1}m.$ What is the ratio of work done by gas in moving the piston from initial position to A , from there to B.

Solution:
The work done till A is given as poAd=105×2×10−2×2×10−1=400J$p_{o}Ad={10}^5\times 2\times {10}^{-2}\times 2\times {10}^{-1}=400J$
The work done from A to B is given as poAd+12kd2=400+12×2000×4×10−2=440J$p_{o}Ad+{\displaystyle \frac{1}{2}}k{d}^2=400+{\displaystyle \frac{1}{2}}\times 2000\times 4\times {10}^{-2}=440J$
Thus we get the ratio as 400440=1011${\displaystyle \frac{400}{440}}={\displaystyle \frac{10}{11}}$

Example

Isothermal piston cylinder

Example: Two cylinders A and B fitted with pistons contain an equal number of moles of an ideal mono-atomic gas at 400K$400K$. The piston of A is free to move while that of B is held fixed Same amount of heat energy is given to the gas in each cylinder. If the rise in temperature of the gas in A is 42K$42K$, the rise in temperature of the gas in B is: (γ=53)$(\gamma ={\displaystyle \frac{5}{3}})$.

Solution:Piston of A is free to move, implies heat is added at constant pressure.
dq=nCpdT1=nCvdT1+nRdT1$dq=n{C}_{p}d{T}_1=n{C}_{v}d{T}_1+nRd{T}_1$

Piston B, the heat is added at constant volume
dq=du=nCvdT2$dq=du=n{C}_{v}d{T}_2$

Given that same amount of heat is supplied.

∴nCvdT1+nRdT1=nCvdT2$\therefore n{C}_{v}d{T}_1+nRd{T}_1=n{C}_{v}d{T}_2$

Givenγ=5/3$\gamma =5/3$⇒Cp=5R/2;Cv=3R/2$\Rightarrow C_p=5R/2;C_v=3R/2$

3R2dT1+RdT1=3R2dT2${\displaystyle \frac{3R}{2}d{T}_1+Rd{T}_1=\frac{3R}{2}d{T}_2}$

dT2=5dT13=53(42)=70K${\displaystyle d{T}_2=\frac{5d{T}_1}{3}=\frac{5}{3}(42)=70K}$

Definition

Calculate work done by a gas in a cylinder where piston is attached to a spring

An unstretched spring is attached to a horizontal, frictionless piston.  Energy is added to the gas inside the cylinder until the pressure in the cylinder is 400kPa$400kPa$.  Determine the work done by the gas on the piston. (Use Patm=75kPa$P_{atm}=75kPa$).
Key Note : The key to solving this problem is to determine the slope and intercept for the linear relationship between the force exerted by the spring on the piston and the pressure within the gas.  This relationship is linear because the pressure within the cylinder is atmospheric pressure plus the spring force divided by the cross-sectional area of the piston.
Given P2=400kPa$P_2=400kPa$    DPiston=0.050m$D_{Piston}=0.050m$
          Patm=75kPa$P_{atm}=75kPa$    k=1kN/m$k=1kN/m$ W=??kJ$W=??kJ$
Assumptions:-
1. The air in the cylinder is a closed system.2. The process occurs slowly enough that it is a quasi-equilibrium process.
3. There is no friction between the piston and the cylinder wall.
4. The spring force varies linearly with position.
Solution 
For a quasi-equilibrium process, boundary or P-V work is defined by : 
W=x1=0∫x2Ftotaldx=x1=0∫x2(Fspring+Fatm)dx$W{=}_{x_1=0}{\int }^{x_2}{F}_{total}dx{=}_{x_1=0}{\int }^{x_2}(F_{spring}+F_{atm})dx$....... (1)
On solving W=kx22+Fatm=kx222+Fatmx2$W={\displaystyle \frac{k{x}^2}{2}}+F_{atm}={\displaystyle \frac{k{x}_2^2}{2}}+F_{atm}{x}_2$.........(2)

Finally, substitute to evaluate the work done by the gas in the cylinder on its surroundings during this process :
W=0.298kJ$W=0.298kJ$

Law

First Law for adiabatic process

For an adiabatic process ΔQ=0$\mathrm{\Delta }Q=0$
Then according to the first law of thermodynamics:
ΔU+ΔW=ΔQ=0$\mathrm{\Delta }U+\mathrm{\Delta }W=\mathrm{\Delta }Q=0$
where Q$Q$ is the heat supplied to the system, W$W$ is the work done by the system and U$U$ is the internal energy of the system.

Definition

Adiabatic free expansion

For adiabatic free expansion of an ideal gas, it is contained in an insulated container and then allowed to expand in vacuum. As there is no external pressure for the gas to expand against, the work done by or on the system is zero. Since this process does not involve any heat transfer or work, therefore, from the First Law of Thermodynamics we can say that the net internal energy change of the system is zero.
i.e ΔU=0$\mathrm{\Delta }U=0$

Law

Differential form of First law of Thermodynamics

δU=δQ−δW$\delta U=\delta Q-\delta W$
where W$W$ is the work done;
U$U$ is the internal energy of the system and 
Q$Q$ is the heat supplied.

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