Thermodynamics Concept Page - 4

Definition

Qualitative explanation of First Law of thermodynamics

The internal energy U$U$ of a system can change through two modes of energy transfer: heat and work. Let
δQ=$\delta Q=$ Heat supplied to the system by the surroundings
δW=$\delta W=$ Work done by the system on the surroundings
δU=$\delta U=$ Change in the internal energy of the system.
The general principle of conservation of energy implies that
ΔQ=ΔU+δW$\mathrm{\Delta }Q=\mathrm{\Delta }U+\delta W$
i.e. the energy (δQ$\delta Q$) supplied to the system goes in partly to increase the internal energy of the system (δ U$\delta U$) and the rest in work on the environment (δW$\delta W$). This is known as the First Law of Thermodynamics. It is simply the general law of conservation of energy applied to any system in which the energy transfer from or to the surroundings is taken into account.

Definition

Sign convention

• When work is done by the system, ΔW$\mathrm{\Delta }W$ is positive.
• When work is done on the system, ΔW$\mathrm{\Delta }W$ is negative.
• When heat is given to the system, ΔQ$\mathrm{\Delta }Q$ is positive.
• When heat is given by the system, ΔQ$\mathrm{\Delta }Q$ is negative.

Definition

Physical situation involving First Law of Thermodynamics

We heat a gas, it expands against a weight, some force (pressure times area) is applied over a distance, and work is done. The change in energy of the system shows the connection between the heat added and work done. 

Example

Heat required in a process

Example: The molar specific heat of hydrogen at constant volume is 5cal/K/mol$5cal/K/mol$ . Find heat required to raise the temperature of 1 gm$1gm$ H2$H_2$ gas by 10 K$10K$ at constant volume.

Solution:
We know that, ΔQ=nCvΔT$\mathrm{\Delta }Q=n{C}_v\mathrm{\Delta }T$ 
Here,ΔT=10K,n=mM=12=0.5,Cv=5cal/mol/K$\mathrm{\Delta }T=10K,n={\displaystyle \frac{m}{M}}={\displaystyle \frac{1}{2}}=0.5,C_v=5cal/mol/K$
Putting these values in above equation we get,
 ΔQ=25cal$\mathrm{\Delta }Q=25cal$

Definition

Isolated system

 An isolated system is a thermodynamic system  enclosed by rigid immovable walls through which neither matter nor energy can pass.
Then according to First Law of Thermodynamics,
ΔU=ΔQ$\mathrm{\Delta }U=\mathrm{\Delta }Q$      (∵ΔW=0$\because \mathrm{\Delta }W=0$ for an isolated system)

Definition

Cyclic Process

 In a cyclic process, the system starts in a particular state and returns to that state after undergoing a few different processes.
From the first law of thermodynamics,
ΔQ=ΔU+ΔW$\mathrm{\Delta }Q=\mathrm{\Delta }U+\mathrm{\Delta }W$
∵ΔU=0$\because \mathrm{\Delta }U=0$ because there is no change in internal energy in a cyclic process.
∴ΔQ=ΔW$\therefore \mathrm{\Delta }Q=\mathrm{\Delta }W$

Diagram

Plot of PV, TV and PT diagrams

Example

Solve problems using first law where process is defined using PV-plot

Example: One mole of diatomic ideal gas undergoes a cyclic process ABC as shown in figure. The process BC is adiabatic. The temperatures at A, B and C
are 400K,800 K$400K,800K$ and 600 K$600K$ respectively. Then find the change in internal energy in the process BC.

Solution:
For a diatomic gas,
Cv=52R$C_v={\displaystyle \frac{5}{2}}R$
ΔUBA=nCV(TB−TA)=1×5R2(800−400)=1000R$\mathrm{\Delta }{U}_{BA}=n{C}_V(T_B-T_A)=1\times {\displaystyle \frac{5R}{2}}(800-400)=1000R$
ΔUAC=ΔQAC−WAC=nCP(TA−TC)−nR(TA−TC)$\mathrm{\Delta }{U}_{AC}=\mathrm{\Delta }{Q}_{AC}-W_{AC}=n{C}_P(T_A-T_C)-nR(T_A-T_C)$
=nCV(TA−TC)=1×5R2(400−600)=−500R$=n{C}_V(T_A-T_C)=1\times {\displaystyle \frac{5R}{2}}(400-600)=-500R$ 

Thus adding up the change in internal energy in both these processes we get change in internal energy from C to B as 500 R$500R$.
As the change in internal energy is a point function we get UBC=−500R

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