A pneumatic mountain bike shock absorber contains a pressurized chamber of nitrogen gas. During a sudden compression, the nitrogen is compressed adiabatically, absorbing the energy of the impact to protect the rider.
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 hard landing, the nitrogen gas in the shock absorber is compressed adiabatically. The work done on the gas during this compression is 62.0 J62.0\text{ J}62.0 J.
Which row in the table below correctly represents the values of WWW, QQQ, and ΔU\Delta UΔU for this compression?
| W / JW \text{ / J}W / J | Q / JQ \text{ / J}Q / J | ΔU / J\Delta U \text{ / J}ΔU / J | |
|---|---|---|---|
| Row A | 62.062.062.0 | 62.062.062.0 | 000 |
| Row B | −62.0-62.0−62.0 | 000 | 62.062.062.0 |
| Row C | 62.062.062.0 | 000 | −62.0-62.0−62.0 |
| Row D | −62.0-62.0−62.0 | −62.0-62.0−62.0 | 000 |
The initial conditions for the nitrogen gas in the chamber are:
During a hard landing, the gas is compressed adiabatically to a volume of 1.50×10−5 m31.50 \times 10^{-5}\text{ m}^31.50×10−5 m3.
Calculate the pressure and temperature of the nitrogen immediately after the compression. Take γ\gammaγ for nitrogen as 1.401.401.40.
The compression described in Part 3 occurs very rapidly. A student suggests that if the rider landed extremely slowly, so that the gas was compressed to the same final volume of 1.50×10−5 m31.50 \times 10^{-5}\text{ m}^31.50×10−5 m3 slowly, the work done to compress the gas would be greater than 62.0 J62.0\text{ J}62.0 J.
Deduce, without calculation, whether the student is correct.
