Thermodynamic Processes Practice Test: Test Your Physics Skills

An adiabatic process is defined by no heat transfer between the system and surroundings, so q=0. Energy can still change due to work, which then must come from (or go into) internal energy.

In a rigid container, volume does not change, so Δv=0. Since pv work is w=∫p dv, no volume change means w=0.

In a rigid container, the gas cannot do pv work because volume is constant. Therefore Δu=q, meaning all the heat added increases internal energy.

For an ideal gas, internal energy is primarily a function of temperature. If temperature stays constant, the change in internal energy is approximately zero.

In an isothermal ideal-gas process, Δu≈0 so 0=q−w. Therefore q=w, and for an expansion with w=+500 J, heat must enter: q=+500 J.

Compression means the gas has Δv<0, so work done by the gas is negative (w<0). To keep temperature constant for an ideal gas (Δu≈0), heat must leave the gas, making q<0.

Apply Δu=q−w with q=2000 J and w=800 J. Δu=2000−800=1200 J, so internal energy increases by 1200 J.

A and B are true by definition of the sign convention. D follows directly from Δu=q−w: if q=0 and w>0, then Δu is negative; C is not always true because Δu and temperature don’t have a universal one-to-one link for all substances and situations.

Rearranging Δu=q−w gives q=Δu+w. Substituting gives q=-250+400=150 J, meaning net heat enters even though internal energy drops because work done is larger.

The first law states Δu=q−w. If q equals w, the difference is zero, so internal energy does not change.

Heat and work describe energy crossing the system boundary due to temperature difference or mechanical interaction. Internal energy u is the stored energy in the system and is a state function.

Using Δu=q−w with compression. Absorbing heat gives q=+900 J, and 'work done on the system' corresponds to w=-300 J. So Δu=900−(−300)=1200 J, meaning internal energy rises.

Insulated means q=0, and compression means work done by the system is negative (w<0). Then Δu=q−w=0−(negative)>0, so internal energy increases.

If w=0, then Δu=q. Releasing heat means q is negative, so Δu=-600 J.

Declaring the sign convention prevents mistakes with plus/minus signs. Checking volume change and insulation tells you whether w or q is zero, which simplifies the first law setup.

Rearrange Δu=q−w to w=q−Δu. With q=300 J and Δu=+50 J, w=300−50=250 J.

Start with 80=−120−w. Adding 120 to both sides gives 200=−w, so w=−200 J, meaning work was done on the system.

Internal energy changes when energy enters or leaves as heat or when work is done. Heat and work are process quantities (energy transfer), while u is stored energy.

Cooling suggests heat is leaving the system (q<0) and doing work means w>0. Both effects tend to reduce internal energy, so Δu is likely negative.

Cooling means heat leaves, so q<0. Compression means work done by the system is negative (w<0), so both can be negative at the same time.