Question

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.

Answer 1

Explanation:

The given data  is as follows.

Mass of oxygen present = 100 mg = $100 \times 10^{-3}$ g

So, moles of oxygen present are calculated as follows.

      n = $\frac{100 \times 10^{-3}}{32}$

         = $3.125 \times 10^{-3}$ moles

Diameter of cylinder = 6 cm = $6 \times 10^{-2}$ m

                              = 0.06 m

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

    A = $\pi \times \frac{(0.06)^{2}}{4}$

        = $2.82 \times 10^{-3} m^{2}$

Length of tube = 11 cm = 0.11 m

Hence, volume (V) = $2.82 \times 10^{-3} \times 0.11$

                              = $3.11 \times 10^{-4} m^{3}$

Now, we assume that the inside pressure is P .

And,   $P_{atm}$ = 100 kPa = 100000 Pa,

Pressure difference = 100000 - P

Hence, force required to open is as follows.

      Force = Pressure difference × A

                = $(100000 - P) \times 2.82 \times 10^{-3}$

We are given that force is 173 N.

Thus,

         $(100000 - P) \times 2.82 \times 10^{-3}$ = 173

Solving we get,

          P = $3.8650 \times 10^{4} Pa$

            = 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

    $38.65 \times 3.11 \times 10^{-4} = 3.125 \times 10^{-3} \times 8.314 \times T$

                    T = 462.66 K

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