Two gases A and B are separated by a partition within a box. Gas A contract having 730j of work done on it . 140 of energy are transfered to gas B from A due to a temperature difference. By how much does the internal energy of A change?
To solve this problem, we need to consider the first law of thermodynamics:
ΔU = Q - W
Where ΔU is the change in internal energy, Q is the heat transferred, and W is the work done on the system.
In this case, gas A contracts and has 730 J of work done on it, so W = 730 J. Additionally, 140 J of energy are transferred from gas A to gas B due to a temperature difference, so Q = -140 J (negative because energy is transferred from A to B).
Plugging these values into the first law of thermodynamics equation, we have:
ΔU = -140 J - 730 J
ΔU = -870 J
Therefore, the internal energy of gas A decreases by 870 J.
To find the change in internal energy of gas A, we need to consider the work done on it and the energy transferred to gas B.
The change in internal energy (ΔU) is given by the formula:
ΔU = Q - W
Where:
- ΔU is the change in internal energy,
- Q is the heat added to the system,
- W is the work done on the system.
In this case, we know that 730 J of work is done on gas A (W = 730 J). We are also told that 140 J of energy is transferred from gas A to gas B (Q = -140 J, since the energy is leaving gas A).
Substituting the known values into the equation, we have:
ΔU = -140 J - 730 J
Simplifying, we get:
ΔU = -870 J
Therefore, the internal energy of gas A decreases by 870 J.