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.

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.

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.

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.

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.

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.

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.

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.

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.

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.

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

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.

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