The air within a piston equipped with a cylinder absorbs 570J of heat and expands from an initial volume of 0.12L to a final volume of 0.88L against an external pressure of 1.0 atm. What is the change in internal energy of the air within the piston?
deltaE=q+w
q=570J
w=-PdeltaV
1 atm = 1.01235 x 10^5 Pa
1,000 L = 1 m^3
.76L*1.01235*10^5Pa=76938.6
that is the answer i got but it said it was wrong and idk why
I don't see that you used 570J anywhere. 0.76 L*atm is 77.01 J on my calculator.
To find the change in internal energy of the air within the piston, we can use the equation:
ΔU = Q - W
Where:
ΔU = change in internal energy
Q = heat absorbed by the system
W = work done on or by the system
In this case, the heat absorbed by the system (Q) is given as 570 J.
To calculate the work done by the system (W), we can use the equation:
W = PΔV
Where:
P = external pressure
ΔV = change in volume
The external pressure (P) is given as 1.0 atm.
The change in volume (ΔV) can be calculated as the difference between the final volume and the initial volume:
ΔV = Vf - Vi
= 0.88 L - 0.12 L
= 0.76 L
Now we can substitute the values into the equation to find the work done:
W = PΔV
= (1.0 atm)(0.76 L)
= 0.76 atm·L
Finally, we can substitute the values of heat (Q) and work (W) into the equation for change in internal energy (ΔU):
ΔU = Q - W
= 570 J - 0.76 atm·L
Note that the units for heat (Q) and work (W) are different, so we need to convert one of them into the same unit before calculating the change in internal energy. Since 1 atm·L = 101.3 J, we can convert work from atm·L to J:
0.76 atm·L = 0.76 atm·L * 101.3 J / 1 atm·L
= 76.888 J
Therefore, we have:
ΔU = 570 J - 76.888 J
= 493.112 J
The change in internal energy of the air within the piston is 493.112 J.