#1 The internal energy of the gas in a gasoline engine’s cylinder decreases by 295 J. If 47.0 J of
work is done by the gas, how much energy is transferred as heat?
#2 If an ideal gas does work on its surroundings:
a. The internal energy of the gas increases
b. The volume of the gas decreases
c. The volume of the gas increases
d. The pressure of the gas increases
Btw , This Is Not Homework This Is A Study Guide For The Final Exam.
#1 change in U = Qin - Wout
-295 = Qin -47
Qin = -248 J, which means that 248 J heat was trasfered OUT.
#2. Work done by a gas = P*(increase in V)
What does that tell you?
Is that your last question for today or forever?
So , It's An increase in pressure or volume ?
And Just Today Because im doing a final review packet.
#1 To find the amount of energy transferred as heat, we can use the First Law of Thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system.
Given:
Change in internal energy (ΔU) = -295 J (decreases by 295 J)
Work done by the gas (W) = 47.0 J
Using the First Law of Thermodynamics:
ΔU = Q - W
Where:
ΔU is the change in internal energy,
Q is the heat added to the system, and
W is the work done by the system.
Plugging in the values:
-295 J = Q - 47.0 J
Solving for Q:
Q = -295 J + 47.0 J
Q = -248 J
Therefore, the energy transferred as heat is -248 J (negative value indicates it is released or transferred out of the system).
#2 If an ideal gas does work on its surroundings, the correct answer is (c) The volume of the gas increases. When a gas does work on its surroundings, it expands, which means the volume of the gas increases. This expansion allows the gas to exert force on the surroundings and do work on them. The internal energy of the gas may or may not increase, depending on the specifics of the system and process. The pressure of the gas can change depending on the volume and temperature of the gas.
#1 To find the energy transferred as heat, we can use the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system.
The formula for the first law of thermodynamics is: ΔU = Q - W
Where:
ΔU is the change in internal energy
Q is the heat added to the system
W is the work done by the system
Given that ΔU = -295 J (decreases by 295 J) and W = 47.0 J (work done by the gas), we can substitute these values into the formula to solve for Q (energy transferred as heat):
-295 J = Q - 47.0 J
Now, we can solve for Q:
Q = -295 J + 47.0 J
Q = -248 J
Therefore, the energy transferred as heat is -248 J (negative sign indicates energy lost).
#2 If an ideal gas does work on its surroundings, the correct option would be:
c. The volume of the gas increases.
When an ideal gas does work on its surroundings, it expands, which causes an increase in volume. As the gas expands, it pushes against the external pressure, doing work in the process. This work is done by the gas on the surroundings and leads to an increase in volume.