How to calculate the area of triangle in pressure volume graphs in thermodynamics?
Answers
An ideal gas in a sealed container is taken through the process shown in the PV diagram below.
Select the correct statement about the signs of the following quantities: change in internal energy of the gas \Delta UΔU, net work done on the gas W,W, and net heat that enters the gas QQ
Choose 1 answer:
Choose 1 answer:
(Choice A)
A
\Delta UΔU is negative, WW is negative, QQ is positive
(Choice B)
B
\Delta UΔU is negative, WW is positive, QQ is negative
(Choice C)
C
\Delta UΔU is negative, WW is positive, QQ is positive
(Choice D)
D
\Delta UΔU is positive, WW is negative, QQ is positive
[What is the solution?]
P \times VPV\Delta U
W
Q
Q=\Delta U-W
\Delta UW
Q=(-)-(+)=-
Q
Example 2: Finding area
An ideal gas in a sealed container is taken through the process shown in the PV diagram below. The initial volume of the gas is V_i=0.25 \text{m}^3V
i
=0.25m
3
and the final volume of the gas is V_f=0.75 \text{m}^3V
f
=0.75m
3
. The initial pressure of the gas is P_i=70,000 \text{ Pa}P
i
=70,000 Pa and the final pressure of the gas is P_f=160,000 \text{ Pa}P
f
=160,000 Pa.
What is the work done on the gas during the process shown?
Solution:
We can find the work done by determining the total area under the curve on a PV diagram. We have to make sure we use the total area, all the way down to the volume axis. For instance, we can imagine viewing the area under the curve in the example shown above as a triangle and a rectangle (as seen below).
Now we just find the sum of the areas of the triangle and rectangle. The height of the rectangle is the pressure P_iP
i
and the width of the rectangle is the change in volume \Delta V=V_f-V_iΔV=V
f
−V
i
. So,
\blueD{\text{area 1}}=\text{height} \times \text{width} \quadarea 1=height×width (area of the rectangle)
\blueD{\text{area 1}}=P_i \times \Delta V \quadarea 1=P
i
×ΔV (height is P_iP
i
and width is \Delta VΔV)
\blueD{\text{area 1}}=(70,000\text{ Pa}) \times (0.75\text{m}^3-0.25\text{m}^3)\quadarea 1=(70,000 Pa)×(0.75m
3
−0.25m
3
) (plug in values)
\blueD{\text{area 1}}=35,000 \text{ J} \quadarea 1=35,000 J (calculate)
We can find the area of a triangle by using A=\dfrac{1}{2}bhA=
2
1
bh.
\greenD{\text{area 2}}=\dfrac{1}{2}bh \quadarea 2=
2
1
bh (area of the triangle)
\greenD{\text{area 2}}=\dfrac{1}{2}b(160,000\text{Pa}-70,000\text{ Pa}) \quadarea 2=
2
1
b(160,000Pa−70,000 Pa) (the height of the triangle is the difference in pressures P_f-P_iP
f
−P
i
)
\greenD{\text{area 2}}=\dfrac{1}{2} (0.75\text{m}^3-0.25\text{m}^3)(160,000\text{Pa}-70,000\text{ Pa}) \quadarea 2=
2
1
(0.75m
3
−0.25m
3
)(160,000Pa−70,000 Pa) (the base of the triangle is the difference in volumes V_f-V_iV
f
−V
i
)
\greenD{\text{area 2}}=22,500 \text{ J} \quadarea 2=22,500 J (calculate)
So the total area under the curve is 35,000 \text{ J} + 22,500 \text{ J}=57,500 \text{ J}35,000 J+22,500 J=57,500 J
This area represents the absolute value of the total work done during the process. To determine the sign of the work done on the gas we notice that the process moves the state to the right, causing the gas to expand. When gas expands the work done on the gas is negative. So,
W_\text{on gas}=-57,500 \text{ J}\quadW
on gas
=−57,500 J