My assignment is as follows:

A sample of air is trapped in a container at $\pu{16.2 psi}$. If the initial temperature is increased from $\pu{10 ^\circ C}$ to $\pu{32 ^\circ C}$, what is the final pressure?

I am really confused as to what $\pu{psi}$ is and how am I supposed to convert it to figure out the final pressure.

  • 1
    $\begingroup$ Strictly speaking, you don't even need to know what psi is. Just apply the ideal gas law and return your answer in the same units. $\endgroup$ Dec 1 '16 at 9:21
  • $\begingroup$ Be careful here, the constant R has many values, some of which works with psi. But yea, apply ideal gas law with the appropriate R and you don't need to concern yourself with psi. R is approximately 5.961 ft3 psi K^-1 lb-mol^-1 $\endgroup$ Dec 1 '16 at 12:16
  • $\begingroup$ Another question is "What does the number 16.2 mean?" Is this the reading on a really accurate (automobile) tire pressure gauge, and is this supposed to help you predict your tire pressure when the temperature rises from 50F to 90F? If so, add 15 (psi) to the 16.2 psi in order to get the absolute pressure from the gauge pressure; then do the calculation. I love it when teachers connect the theoretical world to the practical world. $\endgroup$ Sep 14 '20 at 13:31

In the given experiment, there is no indication that the volume of the container changes and/or a reaction takes place. Both the volume and the amount of substance can therefore assumed to be constant.

Let's assume that the enclosed air can be treated as an ideal gas.

Under these conditions, the pressure is proportional to the absolute temperature (in Kelvin): \begin{align} p &\propto T & \frac{p_1}{p_2} &= \frac{T_1}{T_2} \end{align}

This relation is known as Amonton's Law.

I am really confused as to what psi is and how am I supposed to convert it to figure out the final pressure.

Using the equation above, you would only have to convert the temperatures to Kelvin and could give the final pressure as $\pu{psi}$ too.

If you insist on conversion to Pascal ($\pu{Pa}$) as your favourite SI unit for pressure, remember that a pressure is defined as a force acting on a surface (area): $$\pu{1 Pa} = \pu{1 N//m^2} = \pu{1 kg//m s^2}$$

The $\pu{psi}$ is a still rather common unit and it stands for pounds per square inch.

This is double weird, since both the force and the area are expressed in non-SI units.

Note that the pound is not the German Pfund ($= \pu{0.500 kg}$), but the angloamerican pound with $\pu{1 lb} = \pu{0.453 kg}$.

The area part in $\pu{psi}$ is given in square inches with $\pu{1 in} = \pu{0.0254 m}$.

In summary, $\pu{1 psi} \approx \pu{6895 Pa}$.

  • $\begingroup$ The pound in psi is not pound-mass, but pound-force, the gravitational force on a sample with a mass of a pound on earth at sea level. As long as you are using the Kelvin scale, everything but the psi will cancel out so it does not matter which units everything else is expressed in. $\endgroup$ Sep 14 '20 at 10:32
  • $\begingroup$ @KarstenTheis:(being nitpicky here, but I think it's important to mention) "..expressed in".. until everything else is the same unit (${}^\circ C$ or K) $\endgroup$
    – sai-kartik
    Sep 14 '20 at 13:53

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