Internal energy and the first law of thermodynamics
Internal energy U is the kinetic plus potential energy of the molecules. First law: ΔQ = ΔU + ΔW, which is energy conservation for heat and work.
Internal energy
Water gives out energy when it freezes into ice, so energy must be stored in it. This stored energy is internal energy (U). It is the total kinetic energy of the randomly moving molecules plus their potential energy due to the forces between them. Heating makes the molecules move faster. Doing work on a body, by rubbing or squeezing it, can also raise U. U is a function of state: it depends only on the present state (P, V, T), not on how the system got there. For an ideal gas, U depends only on temperature. So when an ideal gas changes at constant temperature, ΔU = 0. Your NIOS lesson calls ΔU finite in an isothermal process; for an ideal gas it is zero, and the lesson itself uses this for the Carnot cycle.
The first law of thermodynamics
The first law is the law of conservation of energy applied to heat and work. Statement: the heat given to a system equals the increase in its internal energy plus the work done by the system. In symbols, ΔQ = ΔU + ΔW, or ΔU = ΔQ − ΔW. Heat ΔQ and work ΔW both depend on the path taken between two states. But their difference, ΔU, is the same for every path.
Sign convention used in your NIOS lesson
| Quantity | Positive (+) when | Negative (−) when |
|---|---|---|
| Heat ΔQ | Heat is given to the system | Heat is lost by the system |
| Work ΔW | Work is done by the system (it expands) | Work is done on the system (it is compressed) |
| Internal energy ΔU | Internal energy rises (an ideal gas warms up) | Internal energy falls (an ideal gas cools) |
The first law in special processes
| Process | Condition | First law becomes | Meaning |
|---|---|---|---|
| Isothermal (ideal gas) | T constant, so ΔU = 0 | ΔQ = ΔW | All the heat taken in is given out as work |
| Adiabatic | ΔQ = 0 | ΔU = −ΔW | Compression heats the gas; expansion cools it |
| Isochoric | ΔV = 0, so ΔW = 0 | ΔQ = ΔU | All the heat goes into internal energy |
| Cyclic | Back to the start, so ΔU = 0 | ΔQ = ΔW | Net heat taken in = net work done |
Worked example: change in internal energy
Find the change in internal energy of a gas in two cases. (a) It takes in 1500 J of heat and does 400 J of work. (b) It takes in 800 J of heat, and 300 J of work is done on it.
- Use ΔU = ΔQ − ΔW, with the NIOS signs.
- (a) ΔQ = +1500 J, ΔW = +400 J: ΔU = 1500 J − 400 J = 1100 J.
- (b) ΔQ = +800 J, ΔW = −300 J (work done on the gas): ΔU = 800 J − (−300 J) = 1100 J.
In both cases ΔU = +1100 J: the internal energy rises by 1100 J
Limitations of the first law
- It does not tell the direction of heat flow. It would allow heat to flow from a cold body to a hot body by itself, which never happens.
- It does not tell whether heat can change into work. A bullet that hits a target warms it, but the warm target never throws the bullet back.
- It does not tell how much of a given amount of heat can be turned into work.
- These gaps are filled by the second law of thermodynamics.
Rubbing your palms on a cold winter morning raises the internal energy of your skin, so your hands warm up. A car engine turns heat from burning petrol into work, but the first law alone cannot say how much heat becomes work.
3 marks: “Give the first law of thermodynamics in words and as an equation, with the sign convention of your NIOS lesson. Name two things it cannot tell us.” Heat given = rise in internal energy + work done by the system: ΔQ = ΔU + ΔW. Heat in +, work by the system +, rise in U +. Limits: it gives no direction of heat flow, and does not say how much heat can become work.
2 marks: “A gas absorbs 600 J of heat and does 250 J of work. Find the change in its internal energy.” ΔU = ΔQ − ΔW = 600 J − 250 J = 350 J. The internal energy rises by 350 J.
2 marks: “In an adiabatic compression, 150 J of work is done on a gas. Find the change in its internal energy. Does the gas warm or cool?” ΔQ = 0 and ΔW = −150 J, so ΔU = 0 − (−150 J) = +150 J. U rises, so the gas warms up.
Heat in = rise in U + work out: Q = U + W. Work done ON the gas carries a minus sign.