Question #d1389

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Question #d1389

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Answer 1 · 2017-06-01T17:42:26

For ideal gas on a pV diagram, an adiabatic curve is always steeper than an isotherm if they intersect.

Explanation:

The ideal gas equation: $p V = n R T$ $pV = nRT$

$p$ $p$ is pressure
$V$ $V$ is volume
$n$ $n$ is number of moles
$R$ $R$ is ideal gas constant
$T$ $T$ is the absolute temperature

For an isothermal process, the right hand side of the ideal gas equation does not change. Which means

$p V = constant$ $pV = "constant"$

The slope at any point is given by

$\frac{d p}{d V} = - \frac{p}{V}$ ${dp}/{dV} = -p/V$

using implicit differentiation or any other methods.

For adiabatic processes, the path would follow a curve of the following form:

$p V^{γ} = constant$ $pV^gamma = "constant"$

$γ$ $gamma$ in the above equation is treated as a constant and is meant to be the ratio of heat capacities at constant pressure, $c_{p}$ $c_p$ , to heat capacity at constant volume, $c_{V}$ $c_V$ . This can be derived using the first Law of Thermodynamics.

Differetiating to find the slope

$\frac{d p}{d V} = - γ \frac{p}{V}$ ${dp}/{dV} = -gamma p / V$

This looks very similar to the expression for the slope of isotherm, except that gamma is always > 1 and therefore the slope would be steeper.

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Question #d1389

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