1) 60 J of work is done on a gas, and the gas loses 150 J of heat to its surroundings. What is the change in internal energy?
2) A gas starts with 200 J of internal energy. while you add 180 J of heat to the gas, the gas does 70g of work . what is the final internal energy of the gas?
3) While 40 J of work are done on a gas the internal energy goes down by 150 J. what was the value of the heat added to the gas?
2) A gas starts with 200 J of internal energy. while you add 180 J of heat to the gas, the gas does 70g of work . what is the final internal energy of the gas?
3) While 40 J of work are done on a gas the internal energy goes down by 150 J. what was the value of the heat added to the gas?
Answers & Comments
Verified answer
Answer:
1) The first law of thermodynamics 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:
ΔU = Q - W
In this case, 60 J of work is done on the gas, and the gas loses 150 J of heat to its surroundings. So, the change in internal energy is:
ΔU = -150 J - 60 J = -210 J
Therefore, the change in internal energy is -210 J.
2) Using the first law of thermodynamics:
ΔU = Q - W
The initial internal energy of the gas is 200 J. 180 J of heat is added to the gas, and the gas does 70 J of work. So, the change in internal energy is:
ΔU = 180 J - 70 J = 110 J
The final internal energy of the gas is:
Ufinal = Uinitial + ΔU = 200 J + 110 J = 310 J
Therefore, the final internal energy of the gas is 310 J.
3) Using the first law of thermodynamics:
ΔU = Q - W
In this case, 40 J of work is done on the gas, and the internal energy of the gas goes down by 150 J. So, the change in internal energy is:
ΔU = -150 J
The work done by the gas is W = -40 J (since work done on the gas is negative), so the heat added to the gas is:
Q = ΔU + W = -150 J - (-40 J) = -110 J
Therefore, the value of the heat added to the gas is -110 J.