<p>No. After a bit of mathematics, and the inclusion of water vapor, you can also get <span class="math-container">$P=\rho_d R_dT_v$</span>, where <span class="math-container">$\rho_d$</span> is the dry air density and <span class="math-container">$R_d$</span> is the specific gas constant for dry air, and <span class="math-container">$T_v$</span> is the <a href="https://en.wikipedia.org/wiki/Virtual_temperature" rel="nofollow noreferrer">virtual temperature</a>. So isobars are more than just isotherms- they are also dependent on density and water vapor. Check out <a href="https://en.wikipedia.org/wiki/Ideal_gas_law#Molar_form" rel="nofollow noreferrer">Wikipedia's derivation</a> for the proof that <span class="math-container">$PV=nRT$</span> is the same as <span class="math-container">$P=\rho RT$</span>, and then use the fact that <span class="math-container">$\rho$</span> has both dry air and water vapor, to get <span class="math-container">$P=\rho_d R_dT_v$</span>.</p>

CC BY-SA 4.0
2019-12-19T19:46:46.940Z
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.
scorpdaddy
CC BY-SA 4.0
2019-12-20T19:30:17.167Z
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.
BarocliniCplusplus
CC BY-SA 4.0
2019-12-20T19:40:35.507Z
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).
BarocliniCplusplus
CC BY-SA 4.0
2019-12-19T17:10:20.107Z
2019-12-19T17:10:20.107Z
BarocliniCplusplus
<p>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?</p>

CC BY-SA 4.0
2019-12-19T08:35:11.763Z
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.
AtmosphericPrisonEscape
CC BY-SA 4.0
2019-12-19T14:05:57.730Z
So by density, I presume you mean n/V in the equation.  Thanks.
scorpdaddy
CC BY-SA 4.0
2019-12-19T15:22:46.607Z
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.
AtmosphericPrisonEscape
<meteorology><atmosphere>
Does PV=nRT mean that isobars are also isotherms?
scorpdaddy
# Does PV=nRT mean that isobars are also isotherms?

(Asked by: username_0 on 2019-12-18)

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?

**Tags:** meteorology, atmosphere

> **Comments:**
> **username_1**: 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.  
> **username_0**: So by density, I presume you mean n/V in the equation.  Thanks.  
> **username_1**: 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.  

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## 1 answers

### Answer by username_2 on 2019-12-19. Score: 3

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](https://en.wikipedia.org/wiki/Virtual_temperature). So isobars are more than just isotherms- they are also dependent on density and water vapor. Check out [Wikipedia's derivation](https://en.wikipedia.org/wiki/Ideal_gas_law#Molar_form) 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 $P=\rho\_d R\_dT\_v$.

> **Comments:**
> **username_0**: 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.  
> **username_2**: 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.  
> **username_2**: 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).  

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