Understanding the PV Diagram of a Monatomic Gas

In summary, the gas in a PV diagram experiences three processes:1) constant pressure - Q = W2) cycle - W = Q + ΔU3) constant volume - ΔU = Q
  • #1
PhizKid
477
1

Homework Statement


Consider the following PV diagram of one mole of a monatomic gas:

hMivkKO.png


The units are in atm and litres.

Find the change in internal energy, heat, and work for all 3 processes and for the cycle as well.

Homework Equations



PV = nRT
ΔU = Q - W
ΔU = (3/2)*R*ΔT
Q = (5/2)*R*ΔT (for constant pressure)

The Attempt at a Solution



For the work, I just found the areas:

A -> B: 0 J
B -> C: +3.5 J
C -> A: -1 J

For the cycle, the work done is just the total work added:
3.5 - 1 = 2.5 J

For the cycle, the Q is the same as the Work:
2.5 J

For the cycle, the ΔU = 0.

So far I have this (unless I did something wrong):

3535rkr.png


Since for a constant volume process the ΔU = Q, I decided to do that A -> B first.

T_A = (0.5 atm)(1 L) = (1)(0.0821)T, T = 6.09 K
T_B = (0.5 atm)(1 L) = (1)(0.0821)T, T = 36.5 K

ΔT = 30.41 K

ΔU = (3/2)*R*ΔT = (3/2)*0.0821*30.41 = 3.75 J = Q

So:

x45zO8h.png


The constant pressure process looks easier so I'll do C -> A next:

T_C = (0.5 atm)(3 L) = (1)(0.0821)T, T = 18.27 K
T_A = (0.5 atm)(1 L) = (1)(0.0821)T, T = 6.09 K

ΔT = -12.18

Q = (5/2)*R*ΔT = (5/2)*0.0821*(-12.18) = -2.5 J

ΔU = Q - W = (-2.5) - (-1) = -1.5

So:

FW9DwiX.png


Now I can just add up the unknown columns:

ΔU for B -> C: 3.75 + (-1.5) + x = 0, x = -2.25

Q for B -> C: 3.75 + (-2.5) + x = 2.5, x = 1.25

So I got:

ZiCNBTq.png


But I got this problem incorrect. What errors have I made?
 
Last edited:
Physics news on Phys.org
  • #2
Your work looks correct. Are you sure the work asked is the work done by the gas? ehild
 
  • #3
PhizKid said:
But I got this problem incorrect. What errors have I made?
Your method is correct but you have to check your units. In the MKS system the units for pressure are Pascals and the units for volume are m3. Your PV diagram is in atmospheres and litres.

So, for example, the work done by the gas from B to C is not 3.5 Joules but 3.5 atm. litres = 3.5(101325Pa) 10-3 m3 =350 J.

AM
 
  • #4
AM is right. You used litres and atm as units, then you got also the energy, heat, and and work in units (Latm) instead of joules.

ehild
 
  • #5


There are a few errors in your calculations and approach to solving this problem.

1. In your first attempt, you calculated the work done for the cycle (A->B->C->A) as 2.5 J, but this is incorrect. The work done for the entire cycle should be the sum of the work done in each individual process, which would be 0 + 3.5 + (-1) = 2.5 J. This is because work is a path-dependent quantity and cannot be simply added or subtracted for different processes.

2. In your first attempt, you also calculated the heat (Q) for the entire cycle as 2.5 J, which is incorrect. As mentioned above, work is a path-dependent quantity and cannot be added or subtracted for different processes. The correct value for Q for the entire cycle would be the sum of the Q values for each individual process, which would be 3.75 + (-2.5) + (-1.5) = -0.25 J.

3. In your second attempt, you calculated the temperature change for process A->B as 30.41 K, but this is incorrect. The temperature change should be calculated using the ideal gas law, not the specific heat capacity equation. The correct value for the temperature change would be (0.5 atm)(1 L) = (1 mol)(0.0821 L atm/mol K)(ΔT), which gives ΔT = 6.09 K.

4. Similarly, in your second attempt, you calculated the temperature change for process C->A as -12.18 K, but this is incorrect. The correct value should be (0.5 atm)(3 L) = (1 mol)(0.0821 L atm/mol K)(ΔT), which gives ΔT = 18.27 K.

5. In your second attempt, you calculated the heat (Q) for process C->A as -2.5 J, but this is incorrect. The correct value for Q would be (5/2)*(1 mol)(0.0821 L atm/mol K)(18.27 K) = 3.75 J.

6. Finally, in your second attempt, you calculated the change in internal energy (ΔU) for process B->C as -1.5 J, but this is incorrect. The correct value for ΔU would be (3/2)*(
 

What is a PV diagram?

A PV diagram is a graphical representation of the relationship between pressure (P) and volume (V) of a gas. It shows the changes in these variables as the gas undergoes different processes, such as compression and expansion.

What does a PV diagram of a monatomic gas look like?

A PV diagram of a monatomic gas is a straight line when plotted on a Cartesian coordinate system. This is because monatomic gases follow the ideal gas law, which states that the product of pressure and volume is directly proportional to the temperature of the gas.

How can a PV diagram be used to understand the behavior of a monatomic gas?

The shape and position of the PV diagram can give insights into the behavior of a monatomic gas. For example, a horizontal line indicates that the gas is undergoing an isothermal process, where the temperature remains constant. A steeply sloping line indicates a highly compressible gas, while a shallow sloping line indicates a less compressible gas.

What are the different types of processes shown on a PV diagram of a monatomic gas?

The main types of processes shown on a PV diagram of a monatomic gas are isothermal, adiabatic, isobaric, and isochoric processes. An isothermal process occurs at a constant temperature, an adiabatic process occurs without any heat exchange, an isobaric process occurs at a constant pressure, and an isochoric process occurs at a constant volume.

How can the area under a PV diagram be used to calculate the work done by a monatomic gas?

The area under a PV diagram represents the work done by the gas. By calculating the area, we can determine the amount of work done during a particular process. This is useful in understanding the energy changes in a gas and its efficiency in performing work.

Similar threads

  • Introductory Physics Homework Help
Replies
4
Views
819
  • Introductory Physics Homework Help
Replies
1
Views
113
  • Introductory Physics Homework Help
Replies
2
Views
754
  • Introductory Physics Homework Help
Replies
16
Views
897
  • Introductory Physics Homework Help
Replies
4
Views
2K
  • Introductory Physics Homework Help
Replies
6
Views
1K
  • Introductory Physics Homework Help
Replies
3
Views
3K
  • Introductory Physics Homework Help
Replies
12
Views
4K
  • Introductory Physics Homework Help
Replies
11
Views
2K
  • Introductory Physics Homework Help
Replies
4
Views
1K
Back
Top