Internal Energy Temperature Quiz: Test Your Thermodynamics Basics

Concept: Ideal gas internal energy. In the ideal gas model, internal energy is tied to the average kinetic energy of molecules, which depends on temperature. Pressure and volume can change without changing u if temperature stays constant.

Concept: u as a function of t for ideal gases. For an ideal gas, internal energy depends primarily on temperature, so same initial and final temperature implies the same internal energy. Therefore, the change in internal energy is approximately zero.

Concept: Rigid container (w=0) with heat loss. A sealed rigid container means w=0. Releasing heat gives q negative, so Δu=q=−250 J.

Concept: When internal energy change can be zero. A cycle always gives Δu=0 because the system returns to the initial state. For an ideal gas, an isothermal process gives Δu≈0 because temperature is constant, but adiabatic and isochoric processes can still have nonzero Δu.

Concept: Sign logic in Δu=q−w. If q is negative, heat leaves the system, lowering internal energy. Subtracting a positive w lowers it further, so Δu must be negative.

Concept: Work from w=q−Δu. Rearranging gives w=q−Δu. w=1000−300=700 J, meaning the system did 700 J of work.

Concept: Δu is path-independent. Internal energy is a state function, so its change depends only on initial and final states. Different paths can change q and w, but they must combine to give the same Δu.

Concept: Heat/work are process quantities. Heat and work are both energy transfers and are measured in joules. They are not state functions, and many processes have q=0 with w≠0 (adiabatic) or w=0 with q≠0 (isochoric).

Concept: Work done on a system can dominate. If the system releases heat (q<0) but the surroundings do enough work on it (w very negative), then Δu=q−w can still be positive. Compression with some heat loss is a common example.

Concept: Plain-language first law statement. The first law is energy conservation applied to thermal systems. It says internal energy changes when energy is transferred as heat and/or as work across the system boundary.

Concept: Isothermal ideal-gas energy balance. Same temperature implies Δu≈0 for an ideal gas. Then 0=q−w, so w=q=400 J.

Concept: Competing heat loss and compression work. Substitute into Δu=q−w: Δu=−150−(−600)=+450 J. Even though heat leaves, the work done on the gas is larger, so internal energy increases.

Concept: Solving for work from Δu and q. Use Δu=q−w: 200=−500−w. Add 500 to get 700=−w, so w=−700 J (work done on the system).

Concept: Adiabatic expansion causes cooling. Insulated means q=0, and expansion typically means the gas does work (w>0). Then Δu=−w is negative, so internal energy drops and temperature typically falls for gases.

Concept: Same Δu for same endpoints. Path A gives Δu=q−w=900−400=500 J. Path B must have the same Δu, so 500=500−w_b, which implies w_b=0.

Concept: Combining heat loss and work output. Releasing heat means q=−300 and doing work means w=+300. Δu=q−w=−300−300=−600 J, so internal energy decreases by 600 J.

In an isochoric process, the volume remains constant, meaning no work is done on or by the system (w = 0). According to the first law of thermodynamics, the change in internal energy (Δu) is equal to the heat added to the system (q) since there is no work done. Therefore, in this scenario, the work done is zero, and the change in internal energy is solely dependent on the heat transferred.

Concept: Meaning of negative w in this convention. w is defined as work done by the system, so negative w means the surroundings did work on the system. That transfers energy into the system mechanically.

In energy bookkeeping, the convention is that when heat is added to a system, it is considered positive (q > 0), indicating that energy is flowing into the system. Conversely, work done on the system is also viewed as positive (w > 0), meaning energy is being supplied to the system. Thus, the terminology reflects the direction of energy transfer: heat flowing into the system and work being done by the system. This helps maintain clarity in understanding energy changes within thermodynamic processes.

When a gas is compressed significantly, the work done on the gas increases its internal energy. According to the first law of thermodynamics, the change in internal energy (Δu) is equal to the heat added to the system minus the work done by the system. If there is little heat loss during compression, the work input leads to an increase in internal energy, resulting in a positive Δu. This reflects the energy increase due to the compression process, aligning with the expectation of rising temperature and energy in the gas.