-/1 points SerCP10 12 P001. + An ideal gas is enclosed in a cylinder with a mova

Question

-/1 points SerCP10 12 P001. + An ideal gas is enclosed in a cylinder with a movable piston on top of it. The piston has a mass of 8,000 g and an area of 5.00 cm2 and is free to slide up and down, keeping the pressure of the gas constant. (a) How much work is done on the gas as the temperature of 0.105 mol of the gas is raised from 15.0°C to 335°C? (b) What does the sign of your answer to part (a) indicate? O The surroundings do positive work on the gas. The gas does positive work on its surroundings. There is no work done, by the gas or the surroundings.

Explanation / Answer

a) If you do a free-body diagram on the piston, you'd find only two forces acting on it- the force of gravity acting downward, and the force the gas exerts on the piston acting upward. However, initially, the piston is not moving, so
F=ma=0
Fg-Fp=0
Fg=Fp
Recall that the force due to pressure is F=PA where Force = Pressure of the gas times the area it acts on
mg=PA
P=mg/A
P=(8kg)(9.8m/s^2)/(5x10-4m^2) note that all units must be in simplest forms, kg, m^2, Pa, etc
P=1.568x10^5 Pa

We then use the ideal gas law to solve for the initial volume
PV=nRT
(1.568x10^5 Pa) V = (0.105 mol)(8.314 J/mol K) (288K) note to change from celsius to Kelvin, add 273
V=1.6x10^-3 m^3

We know that as the gas is heated, the volume will expand, but the pressure will stay constant, so we use the ideal gas law to find the final volume
PV=nRT
(1.568x10^5 Pa) V = (0.105 mol)(8.314 J/mol K) (608K)
V=3.38 x10^-3 m^3

Now,
W=-PV
W=-(1.568x10^5 Pa)(1.6x10^-3 m^3 - 3.38x10^-3 m^3)
W= 437.63 J

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