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 177 N force to pull the cap off.

What is the temperature of the gas?

Answer 1

Applied Force +Internal Pressure*Area =Atmospheric Pressure*Area

$177N+P_{i}*\pi r^{2}=P_{atm}*\pi r^{2}$

$r=0.06/2=0.03m$

$177N+P_{i}*\pi (0.03m)^{2}=10^{5}Pa*\pi (0.03m)^{2}$

$177N+P_{i}*\pi (0.03m)^{2}=282.74N$

$P_{i}*\pi (0.03m)^{2}=282.74N-177N$

$P_{i}=105.74N/(\pi (0.03m)^{2})$

${\color{Red} P_{i}=3.74*10^{4}Pa}$

=====================

No of moles,$n=0.1g/(32g/mol)$

$n=3.125*10^{-3}mol$

--------------------

Use Formula $PV=nRT$

where V is Volume =$\pi r^{2}h=\pi (0.03m)^{2}(0.11m)$

$T=PV/nR$

$T=3.74*10^{4}Pa*\pi (0.03m)^{2}(0.11m)/(3.125*10^{-3}mol*8.314J/K�mol)$

$T=3.74*10^{4}*\pi 0.03^{2}0.11/(3.125*10^{-3}*8.314)$

ANSWER: $T=447.71K$

======================