Consider a gas in a cylinder at room temperature (T = 293 K), with a volume of 0

Question

Consider a gas in a cylinder at room temperature (T = 293 K), with a volume of 0.125 m3.The gas is confined by a piston with a weight of 150 N and an area of 0.75 m2. Thepressure above the piston is atmospheric pressure.

a. What is the absolute pressure of the gas?

b. How many moles of the gas are there? How many individual gas particles arethere?

c.  The gas is heated, expanding it and moving the piston up. If the volumeoccupied by the gas triples, how much work is done on the gas if the pressure isheld constant during the process?

d. If the walls of the cylinder and the piston itself are good thermal insulators, noheat (Q) flows into or out of the system. Use the first law of thermodynamicsand your answer from part ‘b’ to calculate the change in temperature of the gasin both Kelvin and degrees Celsius.

Explanation / Answer

(a) What is the pressure of the gas?

This can be determined from a free-body diagram of the piston. The weight of the piston acts down, and the atmosphere exerts a downward force as well, coming from force = pressure x area. These two forces are balanced by the upward force coming from the gas pressure. The piston is in equilibrium, so the forces balance. Therefore:

Solving for the pressure of the gas gives:

p = f/a = 150/0.75 = 200 N/m

(C)
An assumption to make here is that the pressure is constant.
Once the gas has expanded, the pressure will certainly be the same as
before because the same free-body diagram applies. As long as the
expansion takes place slowly, it is reasonable to assume that the
pressure is constant.

If the volume has tripled then, and the pressure has remained
the same, the ideal gas law tells us that the temperature must have
tripled too.

The work done by the gas can be determined by working out the
force applied by the gas and calculating the distance. However, the
force applied by the gas is the pressure times the area, so:

W = F s = P A s

and the area multiplied by the distance is a volume,
specifically the change in volume of the gas. So, at constant pressure,
work is just the pressure multiplied by the change in volume:

W = 200 * 0.125 = 25 J

This is positive because the force and the distance moved are in the same direction, so this is work done by the gas.

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