#### Consequences of the First Law, Physics tutorial

Energy Equation:

Energy equations are the equations that express internal energy of the system as a function of variables defining state of the system. Energy equations, like equation of state, are different for different systems or substances. Equation of state and energy equation together totally determine all properties of substance or system. Energy equations are derived independently, but not from equation of state. We are going to consider systems that state can be described by properties P, V. and T.

T and V Independent:

Consider internal energy U as that function of T and V, U(T,V), then change in internal engine dU between two equilibrium states in which temperature and volume differ by dT and dV is

dU = [(∂U/∂T)v]dT + [(∂U/∂V)T]dV

(∂U/∂T)v is slope of isochoric line and (∂U/∂V)T is a slope isothermal line in which U is plotted as the function of T and V. (∂U/∂T)v can be estimated experimentally and it has physical significant.

The first law

dQ = dU + PdV

dQ = (∂U/∂T)vdT + (∂U/∂V)TdV + PdV

Rearranging

dQ = (∂U/∂T)vdT + [(∂U/∂V)T + P]dV

For the process at constant volume, dV = 0 and dQ = CVdT, then equation becomes

CVdT = (∂U/∂T)vdT

CVdT = (∂U/∂T)vdT

So, (∂U/∂T)v = CV

Specific heat capacity at constant volume CV is slope of isochoric line on U-T-V surface and its experimental measurement determines this slope at any point. Equation can be written for any reversible procedure as

dQ = CVdT + [(∂U/∂V)T + P]dV

For the process at constant pressure, dQ = CPdT, so equation becomes

CPdT = CVdT + [(∂U/∂V)T + P]dV

Dividing through by dT and replacing dV/dT by (∂V/∂T)P, we get

CP - CV = [(∂U/∂V)T + P](∂V/∂T)P

Equation holds for the system in any one equilibrium state, but doesn't refer to the process between two equilibrium states.

T and P Independent:

Enthalpy H of the system, like internal energy U, is a property of system which depends on state only and can be stated as the function of any two variables P, V, and T. Each of these relations states the enthalpy surface in rectangular coordinate system in which H is plotted along one axis while other two axes are P and V, P and T, or T and V.

Consider enthalpy as function of T and P i.e. H(T, P) ,

dH = (∂H/∂T)PdT + (∂H/∂P)TdP

From definition of enthalpy for a PVT system:

H = U + PV

Differential of H, dH is

dH = dU + PdV + VdP

Combining equations with first law (i.e. replace dU in equation with dQ - PdV and make dQ the subject) gives

dQ = dH - VdP

Insert 6.8 in 6.11, to obtain

dQ = (∂H/∂T)PdT + [(∂H/∂P)T - V]dP

For the isobaric process ( dP = 0), dQ = CPdT. Therefore

(∂H/∂T)P = CP

Equation means that specific heat capacity at constant pressure CP is equal to slope of isobaric line on H - T - P surface. Equation can be written for any reversible process as

dQ = CPdT + [(∂H/∂P)T - V]dP

In the process at constant volume, dQ = CVdT and

CP - CV = -[(∂H/∂P)T - V](∂P/∂T)V

If temperature is constant, equation becomes

dQ = -[(∂H/∂P)T - V]dP

In the adiabatic process, dQ = 0, then equation becomes

CP(∂T/∂P) = -[(∂H/∂P)T - V]

P and V Independent:

Consider U as the function of P and V, U(P,V), then change in internal energy dU between two equilibrium states is

dU = (∂U/∂P)VdP + (∂U/∂V)PdV

Consider also U(T,V)

dU = (∂U/∂T)VdT + (∂U/∂V)TdV

Generally, for any property w, and any three variables x, y, z the form of equations are

(∂w/∂x)y = (∂w/∂z)y(∂z/∂x)y and (∂w/∂y)x = (∂w/∂z)y(∂z/∂y)x + (∂w/∂y)z

Hence for H(P,V,T) we have

(∂H/∂V)P = (∂H/∂T)P(∂T/∂V)P and

(∂H/∂P)V = (∂H/∂T)P(∂T/∂P)v + (∂H/∂P)T

By solving equations we have

(∂H/∂V)P = CP(∂T/∂V)P

We can show that

dQ = CP(∂T/∂V)pdV + CV(∂T/∂P)vdP and CV(∂P/∂V)s = CP(∂P/∂V)T

Gay-Lussac-Joule Experiment:

The partial derivative (∂U/∂V)T describes way in which internal energy of the given system differs with volume at constant temperature. Likewise (∂H/∂P)T describe way in which enthalpy of the given system differs with pressure at constant temperature. These two derivatives can be computed from equation of state of system. Using

(∂U/∂V)T(∂V/∂T)U(∂T/∂U)v = -1

Then

(∂H/∂P)T(∂P/∂T)H(∂T/∂H)P = -1

Then

(∂H/∂P)T = -Cp(∂T/∂P)H

From equation measurement of rate of change of temperature with volume in the process at constant internal energy provides desired derivatives (i.e. (∂U/∂V)T

Likewise, from equation, measurement of rate of change of temperature with pressure in the process at constant enthalpy provides desired derivatives (i.e. (∂H/∂P)T)

Gay-Lussac and Joule made the attempt to compute dependence of internal energy U of the gas on its volume. In Gay-Lussac and Joule Experimental set-up there is a vessel A that has sample of gas for investigation and is connected to the evacuated vessel B by the tube in which there is stopcock which is at first closed. Whole arrangement is immersed in the water tank of known mass which temperature can be measure by a thermometer. The whole set-up is allowed to attain thermal equilibrium and the temperature is measured and recorded. Then stopcock is opened and gas is permitted to undergo a free expansion into evacuated vessel.

Work done W during free expansion process is zero. The system will finally come to new equilibrium state if pressure is same in both vessels. If temperature of gas changes during process (i.e. free expansion), there will be heat flow between gas and water bath and final temperature will be different from initial one already estimated and recorded.

Gay-Lussac and Joule found that temperature change of water bath, if it changes at all, was very small to be detected. Reason is that heat capacity of bath is so large that small heat flow in or out of it generates only a very small change in temperature. Similar experiments have been carried out, using other method, and results showed that temperature change of gas during free expansion is not large.

Therefore postulate as the additional property of ideal gas is that temperature change during a free expansion is zero. First law of thermodynamics (i.e. ΔU = Uf - Ui = Q - W) as both Q and W are zero, becomes

ΔU = 0

Thus internal energy is constant, and for ideal gas,

(∂T/∂V)U = 0 (ideal gas)

Partial derivative in equation is known as Joule coefficient and is represented by η

η ≡ (∂T/∂V)U

for an ideal gas, the partial derivative (∂U/∂T)v is a total derivative and

Cv = dU/dT, dU = CvdT

Integrating equation from reference level (Uo, To) to (U,T), and if CV is constant that is

U0UdZU = U - U0 = Cv∫TT0dT

That gives

U = U0 + CV(T - T0) This is energy equation of the ideal gas.

Joule-Thompson Experiment:

Joule and Thomson made the attempt to compute dependence of enthalpy of the gas on its pressure. In experimental set-up used by Joule and Thomson gas in compartment 1 (with T1, P1, and V1) was permitted to expand freely through the porous plug. Gas expands from pressure P1 to P2 by throttling action of porous plug. Whole system is insulated so that expansion takes place adiabatically (i.e. Q = 0).

When steady state condition has been reached temperatures of gas before and after expansion, T1 and T2, are estimated directly with sensitive thermocouple thermometers. Total work done during expansion can be written as:

W =W1 +W2 = P1V1 - P2V2

Overall change in internal energy of gas during adiabatic expansion is then

ΔU = Q + W = 0 + W = +W

ΔU = P1V1 - P2V2 = U2 -U1

Rearranging gives

U2 + P2V2 = U1 + P1V1

But H = U + PV then equation becomes H1 + H2

This is hence an isenthalpic expansion and experiment measures directly change in temperature of the gas with pressure at constant enthalpy. Joule-Thomson coefficient μ is,

μ ≡ (∂T/∂P)H

For ideal gas,

(∂H/∂P)T = 0 (ideal gas)

Thus for ideal gas

(∂U/∂V)T = (∂H/∂P)T = 0

Then equation becomes

Cp - Cv = P(∂V/∂T)P = V(∂P/∂T)V = nR

And from equation of state of ideal gas, PV = nRT

P(∂V/∂T)P = V(∂P/∂T)V = nR

Therefore for ideal gas,

Cp - Cv = nR

For any substance in a reversible adiabatic process,

(∂P/∂V)S = CP/CV(∂P/∂V)T

If we representing the ratio CP/CV by γ

That is γ ≡ Cp/CV

Omitting subscript S for simplicity then

dP/P + γdV/V = 0

Integrating equation

ln P + γlnV = lnK

Or PVγ = K

Where K in equation is a constant of integration. From equation eliminating V provides

TP(1-γ)/γ = constant and eliminating P gives

TVγ-1 = constant

Equation based on fact that gas obeys equation of state in any reversible process.

Tutorsglobe: A way to secure high grade in your curriculum (Online Tutoring)

Expand your confidence, grow study skills and improve your grades.

Since 2009, Tutorsglobe has proactively helped millions of students to get better grades in school, college or university and score well in competitive tests with live, one-on-one online tutoring.

Using an advanced developed tutoring system providing little or no wait time, the students are connected on-demand with a tutor at www.tutorsglobe.com. Students work one-on-one, in real-time with a tutor, communicating and studying using a virtual whiteboard technology.  Scientific and mathematical notation, symbols, geometric figures, graphing and freehand drawing can be rendered quickly and easily in the advanced whiteboard.

Free to know our price and packages for online physics tutoring. Chat with us or submit request at info@tutorsglobe.com