00:02
In this problem, we can use newton's second law of motion.
00:07
So for part a, we need to draw the free border diagram of the piston.
00:13
The forces acting on the piston is the weight, which is acting vertically downward.
00:19
And the force from the inside pressure, that will be equivalent to pressure inside into the cross -sectional area of the piston.
00:28
And the downward force from the pressure outside, that will be equivalent to the out, pressure, outward pressure into the cross -sectional area and for circular piston the cross -sectional area is pi r square so for equilibrium we can say the summation of all forces on the piston must be equivalent to zero or can write the inward pressure into the area or pi times r square minus the weight minus the pressure on the outside times pi r square must be equivalent to zero so the pressure on the inside is equivalent to the weight upon pi r square plus the pressure on the outside of the piston now substitute the value for part a to calculate the pressure inside the piston and that will be equivalent to the weight of the piston is uh 36 0 .6 newton upon pi times the radius of the piston is 7 cm or pi times 0 .07 meters square plus the pressure on the outside is 101 .3 into 10 .3 pascal that is the atmospheric pressure.
01:55
So the pressure inside the piston comes out to be 101 .368 into 10 .3 pascals.
02:05
Now part b, using the ideal gas equation, we can write pressure times volume is equivalent to n, r, t, or the number of mole is equivalent to the pressure into the volume upon the gas constant r into the temperature t...