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Rearranging the gas law equation to Pressure = nRT/V shows that a reduction in volume by 10 should yield a 10x increase in pressure.

But compressing air also causes a rise in temperature. Immediately after compression, P should therefore be more than 10x its original value, gradually reducing to 10x as the (closed) system cools down to its original temperature by radiating heat.

Is there a mathematical formula to calculate pressure corrected for temperature increase?

As Gas Law temperature is in Kelvins, how significant will the effect on temperature increase (for 10 to 1 compression) be on the immediate pressure increase?

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What you are describing is the adiabatic process of an ideal gas (i.e. with no or neglectable exchange of heat with the environment). This adiabatic process is described by the state equation $$PV^\gamma=\text{const}$$ where $\gamma$ is the adiabatic index of the gas. Monoatomic gases (i.e. noble gases) have $\gamma=\frac{5}{3}$. Diatomic gases (e.g. nitrogen or oxygen) have $\gamma=\frac{7}{5}$. Air (consisting of 99% diatomic gases) also has approximately $\gamma=\frac{7}{5}$.

Using the ideal gas law $\frac{PV}{T}=\text{const}$, you can further derive adiabatic relations between $P$ and $T$, and likewise between $T$ and $V$. $$P^{1-\gamma}T^\gamma=\text{const}$$ $$TV^{\gamma-1}=\text{const}$$

Applying all these to your example:
When you quickly (i.e. adiabatically) compress the gas volume $V$ by a factor of $\frac{1}{10}$, then the pressure $P$ will increase by a factor of $10^\gamma$, and temperature $T$ will increase by a factor of $10^{1/(\gamma-1)}$.

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  • $\begingroup$ Perfect, thanks so much. $\endgroup$ Commented Apr 14 at 14:05
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This is ideal gas equation only, when you do work it depends in what type of situation you are doing work to reduce volume. Second thing is you are talking about loss of heat that will also take energy out of your system, it depends on situation of work, you will easily find formulas for isothermal, adiabatic conditions.

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