PV diagram of a monatomic ideal gas

[mikefitz]

Homework Statement

Suppose 0.50 mol of a monoatomic ideal gas is changed from state A to state D by one of the processes shown on the PV diagram of the figure below.

http://img382.imageshack.us/img382/784/untitledsc4.png

(This is my edited version of the image)

(a) Find the total work done by the gas if it follows the constant pressure path A-E followed by the constant temperature path E-D.

(b) Calculate the total change in internal energy of the gas during the entire process and the total heat flow into the gas.




I'm having trouble with (a). Is the work done from path A-E, E-D reflected on the first diagram I made, or the second? I guess I'm confused as to whether I'm supposed to use 1 or 2 atm in my calculations. thanks

 

[NTUENG]

Hi Mikefitz,

It has been a while since i touched my thermodynamics but nevertheless i'll try to help.
For part (a),
U need to know that 1 atm = 1.01 x 10^5 Pa and 1 litre = 1000 cm³ = 0.001 m³
Work done from A-E (isobaric) = P dV = 2 x 1.01 x 10^5 Pa x (8-4) x 0.001m³ = 808 J
Work done from E-D (isothermal) = Area of trapezium E-D = 1/2 (2+1) x 1.01 x 10^5 Pa x (16-8) x 0.001m³ = 1212 J
Total work done by gas (expansion) = 808 J + 1212J = 2020 J

For part (b),
U use the 1st law of thermodynamics, dU = Q + W or dU = 3/2nRT to solve for the change in internal energy, dU and heat flow, Q, into the gas.

Hope my A level physics helps. Correct me if I'm wrong.

 

[m2003]

The work done by the gas is represented by the area under the curve on the PV diagram. In this case, the work done will be the sum of the work done along both paths A-E and E-D.

To calculate the work done along path A-E, we can use the formula W = PΔV, where P is the constant pressure (1 atm) and ΔV is the change in volume. From the diagram, we can see that the change in volume along this path is 2 L. Therefore, the work done along path A-E is 1 atm * 2 L = 2 atm*L.

To calculate the work done along path E-D, we need to use the formula W = nRT ln(Vf/Vi), where n is the number of moles (0.50 mol), R is the gas constant (8.314 J/mol*K), T is the temperature (300 K), and Vf and Vi are the final and initial volumes, respectively. From the diagram, we can see that Vf = 4 L and Vi = 2 L. Plugging in these values, we get W = 0.50 mol * 8.314 J/mol*K * 300 K * ln(4/2) = 5.545 J.

Therefore, the total work done by the gas along both paths is 2 atm*L + 5.545 J = 7.545 J.

For (b), the change in internal energy of the gas during the entire process can be calculated using the first law of thermodynamics: ΔU = Q - W, where Q is the heat flow into the gas and W is the work done by the gas. From (a), we know that W = 7.545 J. To calculate Q, we can use the formula Q = nCΔT, where n is the number of moles (0.50 mol), C is the molar specific heat capacity (3/2 R = 12.471 J/mol*K), and ΔT is the change in temperature. Along path A-E, the temperature remains constant at 300 K, so Q = 0. Along path E-D, the change in temperature can be calculated using the ideal gas law: P1V1/T1 = P2V2/T2. Plugging in the values, we get T2 = 450 K. Therefore,

 

[kyle8921]

I would like to clarify that the work done by the gas in this scenario is not reflected on either of the diagrams provided. The PV diagram is used to represent the changes in pressure and volume of the gas, while the work done by the gas is a measure of the energy transferred during the process. Therefore, the work done would depend on the path taken by the gas and the equations used to calculate it.

To answer (a), we can use the formula for work done by a gas, which is given by W = PΔV. In this case, since the gas follows a constant pressure path A-E, we can use the initial pressure of 1 atm and the final volume of 6 L to calculate the work done on this path. This would give us a value of 5 L·atm for the work done on the A-E path.

For the E-D path, we are given the temperature of 300 K, which is constant. This means that the ideal gas law, PV = nRT, can be used to calculate the final volume of the gas on this path. Using the initial pressure of 2 atm and the given temperature, we can find the final volume to be 3 L. Therefore, the work done on the E-D path would be 3 L·atm.

To find the total work done by the gas, we simply add the work done on both paths, giving us a total work of 8 L·atm.

For (b), we can use the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat flow into the system minus the work done by the system. In this case, the change in internal energy would be equal to the heat flow into the gas since the work done on the gas is zero. Using the ideal gas law, we can calculate the initial and final internal energies of the gas in states A and D, respectively. This would give us a change in internal energy of 0.5 mol x (3/2)R x (300 K - 600 K) = -225 J.

To find the total heat flow into the gas, we can use the equation Q = ΔU + W, where W is the total work done by the gas (8 L·atm). This would give us a total heat flow of -225 J + 8 L·atm = -