as a system increases in volume it absorbs 52.5 j of energy in the form of heat form the surroundings. the piston is working against a pressure of 5.00 atm. The final volume of the system is 58,0 L. What is the initial volume of the system if the internal energy of the system decreased by 102.5 J
To solve this problem, we can use the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat absorbed by the system minus the work done by the system:
ΔU = Q - W
In this case, we know that the change in internal energy (ΔU) is -102.5 J (since it decreased), the heat absorbed by the system (Q) is 52.5 J (positive because it is absorbed), and the work done by the system (W) is given by the equation:
W = -PΔV
where P is the pressure and ΔV is the change in volume. In this case, the pressure (P) is 5.00 atm and the final volume (ΔV) is 58.0 L. We need to find the initial volume of the system.
Substituting the given values into the equation for work, we have:
W = -PΔV
W = -(5.00 atm)(58.0 L)
W = -290 atm·L
Now, substituting the known values into the first law of thermodynamics equation, we have:
ΔU = Q - W
-102.5 J = 52.5 J - (-290 atm·L)
Simplifying, we have:
-102.5 J = 52.5 J + 290 atm·L
Next, we subtract 52.5 J from both sides:
-155 J = 290 atm·L
Finally, we divide both sides by 290 atm to solve for the initial volume:
-155 J / 290 atm = L
≈ -0.5345 L
However, volume cannot be negative, so we can discard this value.
Therefore, there is no valid solution for the initial volume of the system given this information.
To solve this problem, we can use the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat absorbed or released by the system plus the work done on or by the system.
The equation for the first law of thermodynamics is:
ΔU = q + w
Where:
ΔU = change in internal energy
q = heat absorbed or released by the system
w = work done on or by the system
In this case, we know that the change in internal energy (ΔU) is -102.5 J (decreased), the heat absorbed (q) is 52.5 J, and the work done (w) is given by the equation:
w = -PΔV
Where:
P = pressure
ΔV = change in volume
In this case, the pressure (P) is 5.00 atm and the change in volume (ΔV) is the final volume (58.0 L) minus the initial volume (Vi). Let's plug in the values we know into the equation for the first law of thermodynamics:
ΔU = q + w
-102.5 J = 52.5 J + (-5.00 atm * (58.0 L - Vi))
Now, let's solve for the initial volume (Vi):
-102.5 J = 52.5 J - 5.00 atm * (58.0 L - Vi)
-102.5 J - 52.5 J = -5.00 atm * (58.0 L - Vi)
-155 J = -5.00 atm * (58.0 L - Vi)
-155 J = -290 atm * L + 5 atm * Vi
155 J = 290 atm * L - 5 atm * Vi
155 J - 290 atm * L = -5 atm * Vi
155 J - 290 atm * L = -5 atm * Vi
Dividing both sides by -5 atm:
(-155 J + 290 atm * L) / -5 atm = Vi
31 L - 58 atm * L/atm = Vi
31 L + 58 L = Vi
Vi = 89 L
Therefore, the initial volume of the system is 89.0 L.