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Chapter 18: Problem 52

Prove that the slope of an adiabat at a given point in a \(p V\) diagram is \(\gamma\) times the slope of the isotherm passing through the same point.

Short Answer

Expert verified
The slope of an adiabat at a given point in a \(pV\) diagram is indeed \(\gamma\) times the slope of the isotherm passing through the same point. This conclusion is drawn from understanding the mathematical expressions for adiabatic and isothermal processes and comparing their slopes.

Step by step solution

01

Define the adiabatic process

In an adiabatic process, the ideal gas law transforms to \(pV^\gamma = \text{constant}\), where \(p\) is pressure, \(V\) is volume and \(\gamma\) is the ratio of specific heats, given as \(C_p / C_v\). Taking the total differential, we get \((\gamma pV^{\gamma - 1})dV + V^\gamma dp = 0\). We can solve for \(dp/ dV\) to get the slope of the adiabat on the \(pV\) diagram: \(\frac{dp}{dV} = -\gamma \frac{p}{V}\).
02

Define the isothermal process

In an isothermal process, the ideal gas law becomes \(pV = \text{constant}\). Again, taking the total differential, we obtain \(V dp + p dV = 0\). Solving for \(dp/ dV\), we find the slope of the isotherm: \( \frac{dp}{dV} = - \frac{p}{V}\).
03

Show the relationship

By comparing the slopes obtained from the adiabatic and isothermal processes, we can see that the slope of an adiabat at a given point in a \(pV\) diagram is \(\gamma\) times higher than the slope of the isotherm passing through the same point. This is because the slope of the adiabat is \(-\gamma p / V\) while the slope of the isotherm is \(-p / V\). Thus, the slope of the adiabat is \(\gamma\) times the slope of the isotherm.

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Most popular questions from this chapter

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.

If an ice cube melts into water at \(0^{\circ} \mathrm{C}\), is there any work done on the system during the change?

A gasoline engine has compression ratio 8.5 (see Example 18.3 for the meaning of this term), and the fuel鈥揳ir mixture compresses adiabatically with $$ \gamma=1.4 $$ If the mixture enters the engine at 34掳C, what will its temperature be at maximum compression?

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. $$ \begin{array}{|l|l|l|l|l|l|l|l|} \hline \begin{array}{l} \text { Volume, } \\ V(\mathrm{~L}) \end{array} & 1.1 & 1.27 & 1.34 & 1.56 & 1.82 & 2.14 & 2.37 \\ \hline \begin{array}{l} \text { Pressure, } \\ p(\mathrm{~atm}) \end{array} & 0.998 & 0.823 & 0.746 & 0.602 & 0.493 & 0.372 & 0.344 \\ \hline \end{array} $$

A gas undergoes an adiabatic compression during which its volume drops to half its original value. If the gas pressure increases by a factor of \(2.51\), what's its specific-heat ratio \(\gamma\) ?
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