Answer:
ΔU = - 310.6 J (negative sign indicates decrease in internal energy)
W = 810.6 J
Explanation:
a.
Using first law of thermodynamics:
Q = ΔU + W
where,
Q = Heat Absorbed = 500 J
ΔU = Change in Internal Energy of Gas = ?
W = Work Done = PΔV =
P = Pressure = 2 atm = 202650 Pa
ΔV = Change in Volume = 10 L - 6 L = 4 L = 0.004 m³
Therefore,
Q = ΔU + PΔV
500 J = ΔU + (202650 Pa)(0.004 m³)
ΔU = 500 J - 810.6 J
<u>ΔU = - 310.6 J (negative sign indicates decrease in internal energy)</u>
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b.
The work done can be simply calculated as:
W = PΔV
W = (202650 Pa)(0.004 m³)
<u>W = 810.6 J</u>
Answer:
The work done on the system is -616 kJ
Explanation:
Given;
Quantity of heat absorbed by the system, Q = 767 kJ
change in the internal energy of the system, ΔU = +151 kJ
Apply the first law of thermodynamics;
ΔU = W + Q
Where;
ΔU is the change in internal energy
W is the work done
Q is the heat gained
W = ΔU - Q
W = 151 - 767
W = -616 kJ (The negative sign indicates that the work is done on the system)
Therefore, the work done on the system is -616 kJ
Answer:
Apply Newton's second law in the moving direction.
Explanation:
Friction force applies in the opposite direction of motion; as a restriction.
Answer:
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Explanation:
