# What is the pressure immediately after compression of air?

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?

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 $$V$$. $$P^{1-\gamma}T^\gamma=\text{const}$$ $$TV^{\gamma-1}=\text{const}$$
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 $$10^{1/(\gamma-1)}$$.