Question

Challenge: pV work 2 Consider a container with a frictionless piston that contains a given amount of CO2. Assume that the behavior of this gas can be described by the van der Waals equation of state. For carbon dioxide gas (CO2), the constants in the van der Waals equation are a=0.364J⋅m3/mol2 and b=4.27×10−5m3/mol. Let’s assume that initially the external pressure is 2.20 bar, which is the sum of a 1 bar atmospheric pressure and the pressure created by a very large number of very small pebbles that rest on top of the piston. The initial volume of gas is 0.2 L  and the initial temperature is 25°C. Now, you will increase the volume of the gas by changing the external pressure slowly in a way that guarantees that the temperature of the system remains constant throughout the process. To do this, imagine you remove the pebbles one by one slowly to increase the volume by an infinitesimal amount. Every time you remove a weight you allow the system to equilibrate. Your cylinder is immersed in a water bath at 25°C, which keeps your gas at the same temperature throughout the whole process. Remember to use three significant figures for all numerical answers. The margin of error for each numerical answer is 1%. To avoid rounding errors use the unrounded intermediate values in your final calculations. Note: You may find an equation to solve this problem in a textbook or online, but the goal of this challenge is that you think through the problem and come up with the equation on your own. This problem requires basic calculus, so be ready to integrate!

Part A What is the volume of the gas when you remove all pebbles? V = L SubmitMy AnswersGive Up Part B What is the final pressure of the gas? p = bar SubmitMy AnswersGive Up Part C Now consider the work performed by the sytem. What is the sign of w? Now consider the work performed by the sytem. What is the sign of w? zero negative positive SubmitMy AnswersGive Up Part D What is the value of w? w = J SubmitMy AnswersGive Up

Answer 1

First, let us model via Van der Waals:

The Van der Waals equation is a description of real gases, it includes all those interactions which we previously ignore in the ideal gas law.

It considers the repulsion and collision, between molecules of gases. They are no longer ignored and they also are not considered a"point" particle.

The idel gas law:

PV = nRT

P(V/n) = RT ; let V/n = v; molar volume

P*v = RT

now, the van der Waals equation corrects pressure and volume

(P+ a/v^2) * (v - b) = RT

where;

R = idel gas law; recommended to use the units of a and b; typically bar/atm and dm/L

T = absolute temperature, in K

v = molar volume, v = Volume of gas / moles of gas

P = pressure of gas

Knowing this data; we can now substitute the data

given

a = 3.6551

b = 0.04281

(P+ a/v^2) * (v - b) = RT

A)

a=0.364 J⋅m3/mol2 and b=4.27×10−5m3/mol.

Pexternal initial = 2.20 bar = 2.2*10^5 Pa Vinitial = 0.2L = 0.2*10^-3 m3; Tinitial = 25°C = 298K Tis constant

Pexternal final = 1 bar (atmospheric after removing all pebbles) = 10^5 Pa; Tfinal = 298 since isothermal; Vfinal

First, get moles initially:

(P+ a/v^2) * (v - b) = RT

Solve for v

Pv -Pb + a/v - ab/v^2 = RT

multiply all by v^2

Pv^3 -Pbv^2 + av - ab = RT*v^2

(P)*v^3 + (-P*b -RT)*v^2 + (a)*v - ab = 0

substitute data

(2.2*10^5)*v^3 + (-(2.2*10^5)(4.27*10^-5) - (8.314*298) )*v^2 + (0.364 )*v - (0.364 )(4.27*10^-5) = 0

(2.2*10^5)*v^3 - 2486.96*v^2 + 0.364*v - 0.0000155428 = 0

sovle fo v

v = 0.011156 m3/mol

This is mlar volume, so V = 0.2 L --> 0.2*10^3 m3

V/n = v

n = V/v = (0.2*10^-3)/(0.011156 ) = 0.01792 moles

Now...

if Pnew = 1 bar = 10^5 Pa, T = 298K remains, and Vfinal = ?

substitute

(P)*v^3 + (-P*b -RT)*v^2 + (a)*v - ab = 0

(10^5)*v^3 + (-(10^5)(4.27*10^-5) -(8.314*298) )*v^2 + (0.364)*v - (0.364*4.27*10^-5) = 0

(10^5)*v^3 + (-2481.84 )*v^2 + (0.364)*v - 0.0000155428 = 0

v = 0.024671 m3/mol

Vtotal = 0.024671 * 0.01792 = 0.000442104 m3

In liters --> 0.000442104*10^3 = 0.442104 L

it goes from 0.2 L to 0.442 Liters

b)

Final pressure on gas will be P = 1 bar, since it is the sae as the external pressure, which at the end will be equal to the atmospheric pressure = 1 bar

P = 1 bar

c)

Since the systm is expanding, it must be exerting energy to the surroundings... Therefore, the work is NEGATIVE

d)

Find W

The PV work fr van der waals is given by -->

P = nRT/(V-nb) - a*n^2 / V^2 --->

W - Integral (P*dV) from V1 to V2; solving

--> -Integral (nRT/(V-nb)*dV + integral(an^2/V^2*dV;

W = -n*R*T*ln((V2-nb)/(V1-nb)) + a*n^2*(1/V1 -1/V2)

now, substitute data

it goes from 0.2 L to 0.442 Liters

W = -n*R*T*ln((V2-nb)/(V1-nb)) + a*n^2*(1/V1 -1/V2)

Get

(V2-nb) = (0.442*10^-3 -(0.01792)(4.27*10^-5)) = 0.00044123

(V1-nb) =  (0.200*10^-3 -(0.01792)(4.27*10^-5)) = 0.0001992348

get

a*n^2*(1/V1 -1/V2)

a*n^2*(1/V1 -1/V2) = (0.364)(0.01792^2)(1/(0.200*10^-3) - 1/(0.442*10^-3)) = 0.319993

now,

W = -n*R*T*ln((V2-nb)/(V1-nb)) + a*n^2*(1/V1 -1/V2)

W = -n*R*T*ln(0.00044123/0.0001992348) + 0.319993

W = -0.01792*8.314*298*ln(2.214623) + 0.319993

W = -35.30 + 0.319993

W = -34.98 J

as expected, sign is negative