An industrial hydraulic cylinder contains a small trapped air bubble. During rapid operation, this air pocket is compressed adiabatically, which can affect the system's performance.
The first law of thermodynamics can be written as:
Q=ΔU+W Q = \Delta U + W Q=ΔU+WState what ΔU\Delta UΔU represents in this equation.
During a sudden movement of the hydraulic actuator, the air bubble is compressed adiabatically. The work done on the air during this compression is 35.0 mJ35.0\text{ mJ}35.0 mJ.
Which row in the table below correctly represents the values of WWW, QQQ, and ΔU\Delta UΔU for this compression?
| W / mJW \text{ / mJ}W / mJ | Q / mJQ \text{ / mJ}Q / mJ | ΔU / mJ\Delta U \text{ / mJ}ΔU / mJ | |
|---|---|---|---|
| Row A | −35.0-35.0−35.0 | 000 | 35.035.035.0 |
| Row B | 35.035.035.0 | 35.035.035.0 | 000 |
| Row C | −35.0-35.0−35.0 | −35.0-35.0−35.0 | 000 |
| Row D | 35.035.035.0 | 000 | −35.0-35.0−35.0 |
The initial conditions for the trapped air are:
During rapid operation, the air bubble is compressed adiabatically to a volume of 5.00×10−8 m35.00 \times 10^{-8}\text{ m}^35.00×10−8 m3.
Calculate the pressure and temperature of the air immediately after the compression. Take γ\gammaγ for air as 1.401.401.40.
To produce the adiabatic change, the actuator moves very quickly. An engineer suggests that by operating the actuator slowly, the work done to compress the bubble to the final volume of 5.00×10−8 m35.00 \times 10^{-8}\text{ m}^35.00×10−8 m3 will be greater than 35.0 mJ35.0\text{ mJ}35.0 mJ.
Deduce, without calculation, whether the engineer is correct.
29 exam-style questions on AQA A Level Physics Thermodynamics and engines (A-level only), covering First law of thermodynamics (A-level only), Non-flow processes (A-level only), The $p$ - $V$ diagram (A-level only), Engine cycles (A-level only), Second Law and engines (A-level only), and Reversed heat engines (A-level only). Each one has a worked solution and a mark scheme showing where the marks go.