A 6.0-cm-diameter, 11-cm-long cylinder contains 100 mg of oxygen (O2) at a pressure less than 1 atm. The cap on one end of the cylinder is held in place only by the pressure of the air. One day when the atmospheric pressure is 100 kPa, it takes a 173 N force to pull the cap off.

Explanation:

The given data  is as follows.

Mass of oxygen present = 100 mg = g

So, moles of oxygen present are calculated as follows.

      n =

         = moles

Diameter of cylinder = 6 cm = m

                              = 0.06 m

Now, we will calculate the cross sectional area (A) as follows.

    A =

        =

Length of tube = 11 cm = 0.11 m

Hence, volume (V) =

                              =

Now, we assume that the inside pressure is P .

And,   = 100 kPa = 100000 Pa,

Pressure difference = 100000 - P

Hence, force required to open is as follows.

      Force = Pressure difference × A

                =

We are given that force is 173 N.

Thus,

          = 173

Solving we get,

          P =

            = 38.65 kPa

According to the ideal gas equation, PV = nRT

So, we will put the values into the above formula as follows.

                PV = nRT

                    T = 462.66 K

Thus, we can conclude that temperature of the gas is 462.66 K.

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