Chapter 18: Problem 7
When you let air out of a tire, the air seems cool. Why? What kind of process is occurring?
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Chapter 18: Problem 7
When you let air out of a tire, the air seems cool. Why? What kind of process is occurring?
These are the key concepts you need to understand to accurately answer the question.
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A gas with \(\gamma=1.40\) occupies \(6.25 \mathrm{~L}\) when it's at \(98.5 \mathrm{kPa}\) pressure. (a) What's the pressure after the gas is compressed adiabatically to \(4.18\) L? (b) How much work does that compression require?
Three identical gas–cylinder systems are compressed from the same initial state to final states that have the same volume, one isothermally, one adiabatically, and one isobarically. Which system has the most work done on it? The least?
The adiabatic lapse rate is the rate at which air cools as it rises and expands adiabatically in the atmosphere (see Application: Smog Alert, on page 352). Express \(d T\) in terms of \(d p\) for an adiabatic process, and use the hydrostatic equation (Equation 15.2) to express \(d p\) in terms of \(d y\). Then, calculate the lapse rate \(d T / d y\). Take air's average molecular weight to be \(29 \mathrm{u}\) and \(\gamma=1.4\), and remember that the altitude \(y\) is the negative of the depth \(h\) in Equation 15.2.
An ideal gas with \(\gamma=1.3\) is initially at \(275 \mathrm{~K}\) and \(100 \mathrm{kPa}\). The gas is compressed adiabatically to \(240-\mathrm{kPa}\) pressure. Find its final temperature.
The table below shows measured values of pressure versus volume for an ideal gas undergoing a thermodynamic process. Make a \(\log -\log\) plot (logarithm of \(p\) versus logarithm of \(V\) ) of these data and use it to determine (a) whether the process is isothermal or adiabatic and (b) the temperature if it's isothermal or the adiabatic exponent \(\gamma\) if it's adiabatic. $$ \begin{array}{|l|l|l|l|l|l|l|l|} \hline \begin{array}{l} \text { Volume, } \\ V(\mathrm{~L}) \end{array} & 1.1 & 1.27 & 1.34 & 1.56 & 1.82 & 2.14 & 2.37 \\ \hline \begin{array}{l} \text { Pressure, } \\ p(\mathrm{~atm}) \end{array} & 0.998 & 0.823 & 0.746 & 0.602 & 0.493 & 0.372 & 0.344 \\ \hline \end{array} $$
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