Wednesday, May 25, 2011

Basic Physics (Thermodynamics) - Pt. 12

Having examined how heat and mechanics are related, and specifically how Newton’s laws give rise to the basic kinetic theory for ideal gases, we now enter the realm of thermodynamics proper. We begin by further inquiring into the link between kinetic theory and temperature, and that we already saw:

PV = 2/3 N(½mv^2)

Which can be compared to what is called the “empirical equation of state for an ideal gas”:

PV = nkT, where k is the Boltzmann constant.

Equating the two:

2/3 N(½mv^2) = N kT

And solving for T, temperature:

T = 2/3k [(½mv^2)]

Which discloses the direct link between temperature and the microscopic behavior of a gas. Thus, temperature T is indeed a direct measure of the average molecular kinetic energy of a gas. We can also write this:

3k T/ 2 = (½mv^2)

Now, the total translational kinetic energy (for all N molecules of a gas) is just:

E = N((½mv^2) = N (3kT/2) = 3 nRT/2

Where we have replaced k using k = R/N_A

Where R is the molar gas constant and N_A is the “Avogradro number”

R = 8.3 J/ mole-K

Now, with these preliminaries out of the way, we can explore further the First Law of Thermodynamics (introduced in the two earlier blogs) but now in terms of the experimental set up shown. Here we have an apparatus consisting of a gas confined by a movable piston such that when heat (Q) is applied (added) using the Bunsen burner, external work W (= P(V2 – V1) ) is done in expanding the contained gas and hence pushing the piston upward. From the First Law:

+Q = +U + delta W or

delta Q = delta U + p(V2 – V1) = p(delta V)

Note again that (delta U) includes translational, rotational and vibrational kinetic molecular energies. The pressure itself P = F(a)/ A where F(a) > mg and A is the area of the piston.

Now, consider the experiment in the context of keeping the pressure P constant, then we say the process is isobaric. We also allow that it is reversible, in other words just as I can increase Q to do work to expand the gas, so also I can reduce Q (by lowering the heat of the burner) to decrease the gas.

Given n moles of an ideal gas (taking n = const.) then we can write:

delta Q = n Cp,m (delta T)

where: C p,m = delta Q/ n (delta T)

is the molar heat capacity at constant pressure

Then for an ideal gas, taking only V and T changing:

PV = nR T -> P(delta V) = nR(delta T)


delta Q = delta U + delta W becomes

n Cp,m (delta T) = n Cv,m (delta T) + nR (delta T)


n Cv,m (delta T) = delta U

Using some algebra on the earlier equation, and canceling out delta T’s:

C p,m = C v,m + R

Or R = C p,m - C v,m

In other words, the molar gas constant R is the difference between the molar specific heat capacity at constant pressure and the molar specific heat capacity at constant volume.

Sample Problem:

One mole of a gas has a volume of 0.0223 cubic meters at a pressure P = 1.01 x 10^5 N/m^2 at 0 degrees Celsius. If the molar heat capacity at constant pressure is 28.5 J/mol-K find the molar heat capacity at constant volume, C v,m.


We have: PV = nRT and

R = PV/T = [(1.01 x 10^5 Pa) (0.0223 m^3)]/ 273 K

Note: that 0C = 273 Kelvin (K) and for pressure, 1 Pa (Pascal) = 1 N/m^2

Then R = 8.3 J/ mol K


C v,m = C p,m – R = (28.5 – 8.3) J/mol K = 20.2 J /mol K

Problem for ambitious readers:

20 g of a gas initially at 27 C is heated at a constant pressure of 101 kPa (kiloPascals), so its volume increases from 0.250 m^3 to 0.375 m^3. Find:

a) the external work done in the expansion

b) the increase in the internal energy U

c) the quantity of heat supplied (Q) to achieve the expansion.

No comments: