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.

b) returns freshwater to the earth.

c) returns freshwater to the earth.c) occurs after condensation.

d) returns freshwater to the earth.c) occurs after condensation.d) roughly equal to the amount of evaporation.

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