Does the ideal gas law, PV=nRT, mean that isobars on a map of surface conditions are also isotherms? Is there a way to intuit isotherms from isobars, and vice versa?
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$\begingroup$ At the surface level probably not due to the presence of friction. But in the free troposphere more likely. atmos.illinois.edu/~snesbitt/ATMS505/stuff/… $\endgroup$– user1066Commented Dec 19, 2019 at 4:58
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$\begingroup$ No, only if you keep the density constant, which is not given. Temperature and density are two independent parameters of a given set of particles. $\endgroup$– AtmosphericPrisonEscapeCommented Dec 19, 2019 at 8:35
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$\begingroup$ So by density, I presume you mean n/V in the equation. Thanks. $\endgroup$– scorpdaddyCommented Dec 19, 2019 at 14:05
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$\begingroup$ You should use @-tags otherwise people don't see you've replied to them. Well, $n/V$ is the number density, another common form of the ideal gas law is $P=\frac{\rho k_B T}{\mu}$, which uses the mass density, which i was referring to, in order to eliminate the volume, which is irrelevant in atmospheric science, as the atmosphere is a continuum. $\endgroup$– AtmosphericPrisonEscapeCommented Dec 19, 2019 at 15:22
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No. After a bit of mathematics, and the inclusion of water vapor, you can also get $P=\rho_d R_dT_v$, where $\rho_d$ is the dry air density and $R_d$ is the specific gas constant for dry air, and $T_v$ is the virtual temperature. So isobars are more than just isotherms- they are also dependent on density and water vapor. Check out Wikipedia's derivation for the proof that $PV=nRT$ is the same as $P=\rho RT$, and then use the fact that $\rho$ has both dry air and water vapor, to get $P=\rho_d R_dT_v$.
answered Dec 19, 2019 at 17:10
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$\begingroup$ So over a large area where the temperature ranges from -16°C to -9°C the impact of the ratio of absolute temperatures has a much smaller impact on the pressure than the "squishedness" of the air. Where air is squished we see high pressure and where is is "stretched" we see low pressure. The isobars more closely ressemble iso-density than isothermal lines. $\endgroup$ Commented Dec 19, 2019 at 19:46
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$\begingroup$ No, it represents both. So, from the $P=\rho R T$, equation, we can derive the equation for the pressure gradient: $\nabla P = R\rho \nabla T + RT\nabla \rho$. So one isobar is the combination of both the overlapping isotherm and isoster, weighted by density and temperature, respectively. Much of this complexity is reduced when we just consider an isobaric chart. $\endgroup$ Commented Dec 20, 2019 at 19:30
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$\begingroup$ So in other words, you need more information than just the temperature range. You also need the pressure overlay (what is the pressure at those locations of maximum and minimum temperature). $\endgroup$ Commented Dec 20, 2019 at 19:40