Relation between CP & CV Recall: Can be related

Relation between CP & CV U is a function of T and V Thus; At constant P;

Relation between CP & CV Dividing by dTP ;

Joule Experiment To determine  - A sample of a gas (the system) was placed in one side of the apparatus and the other side of the apparatus was evacuated. -The initial temperature of the apparatus was measured. -The stopcock was then opened and the gas expanded irreversibly into the vacuum. 

Joule Experiment -Because the surroundings were not affected during the expansion into a vacuum, w was equal to zero. -The gas expanded rapidly so there was little opportunity for heat to be transferred to or from the surroundings. - If a change in temperature of the gas occurred, heat would be transferred to or from the surroundings after the expansion was complete, and the final temperature of the surroundings would differ from the initial temperature.

Joule Experiment To determine Joule coefficient, Joule experiment gave and because the changes in temperature that occurred were too small to be measured by the thermometer.

Joule-Thomson experiment To determine Insulated wall Initial State Final State

Joule-Thomson experiment Adiabatic process Rearranging; Isenthalpic process

Joule-Thomson experiment To determine Joule-Thomson coefficient,

Perfect Gases Perfect gas: one that obeys both of the following: U is not change with V at constant T If we change the volume of an ideal gas (at constant T), we change the distance between the molecules. Since intermolecular forces are zero, this distance change will not affect the internal energy.

Perfect Gases For perfect gas, internal energy can be expressed as a function of temperature (depends only on T): An infinesimal change of internal energy, From (slide 19)

This shows that H depends only on T for a perfect gas Perfect Gases For perfect gas, enthalpy depends only on T; Thus; An infinesimal change of enthalpy, This shows that H depends only on T for a perfect gas From (slide19)

Perfect Gases The relation of CP and CV for perfect gas; From slide 24 PV=nRT, thus or

Perfect Gases For perfect gas, - Since U & H depend only on T.

Perfect gas and First Law First law; For perfect gas; Perfect gas, rev. process, P-V work only

Reversible Isothermal Process in a Perfect Gas First law: Since the process is isothermal, ∆T=0; Thus, ∆U = 0 First law becomes;

Reversible Isothermal Process in a Perfect Gas Since , thus: Work done; rev. isothermal process, perfect gas OR

Reversible Adiabatic Process in Perfect Gas First law, Since the process is adiabatic, For perfect gas, becomes, Since

Reversible Adiabatic Process in Perfect Gas Since , thus By integration;

Reversible Adiabatic Process in Perfect Gas If is constant, becomes; OR rev. adiabatic process, perfect gas

Reversible Adiabatic Process in Perfect Gas Alternative equation can be written instead of Consider

Reversible Adiabatic Process in Perfect Gas Since rev. adiabatic process, perfect gas, CV constant where

Reversible Adiabatic Process in Perfect Gas If is constant, becomes rev. adiabatic process, perfect gas, CV constant

Exercise A cylinder fitted with a frictionless piston contains 3.00 mol of He gas at P = 1.00 atm and is in a large constant- temperature bath at 400 K. The pressure is reversibly increased to 5.00 atm. Find w, q and ∆ U for the process.

Solution A cylinder fitted with a frictionless piston contains 3.00 mol of He gas at P = 1.00 atm and is in a large constant- temperature bath at 400 K. The pressure is reversibly increased to 5.00 atm. Find w, q and ∆ U for the process. Assumption: He as a perfect gas. ∆U =0 since T is constant (U depends only on T)

Solution

Calculation of First Law Quantities 1. Reversible phase change at constant P & T. Phase change: a process in which at least one new phase appear in a system without the occurance of a chemical reaction (eg: melting of ice to liquid water, freezing of ice from an aqueous solution.)

Calculation of First Law Quantities 2. Constant pressure heating with no phase change

Calculation of First Law Quantities 3. Constant volume heating with no phase change Recall:

Calculation of First Law Quantities 4. Perfect gas change of state: H & U of a perfect gas depends on T only;

Calculation of First Law Quantities 5. Reversible isothermal process in perfect gas H & U of a perfect gas depends on T only; Since , thus

Calculation of First Law Quantities 6. Reversible adiabatic process in perfect gas If Cv is constant, the final state of the gas can be found from: where Adiabatic process

Exercise CP,m of a certain substance in the temperature range 250 to 500 K at 1 bar pressure is given by CP,m = b+kT, where b and k are certain known constants. If n moles of this substance is heated from T1 to T2 at 1 bar (where T1 & T2 are in the range of 250 to 500 K), find the expression for ΔH.

SOLUTION CP,m of a certain substance in the temperature range 250 to 500 K at 1 bar pressure is given by CP,m = b+kT, where b and k are certain known constants. If n moles of this substance is heated from T1 to T2 at 1 bar (where T1 & T2 are in the range of 250 to 500 K), find the expression for ΔH. - P is constant throughout the heating, thus;