Suppose a monatomic ideal gas is contained within a vertical cylinder that is fitted with a movable piston. The piston is frictionless and has a negligible mass. The area of the piston is 3.06 10-2 m2, and the pressure outside the cylinder is 1.02 105 Pa. Heat (2109 J) is removed from the gas. Through what distance does the piston drop?
PV=work
V= area *distance
You know P, area, and work. Solve for distance.
To find the distance through which the piston drops, we can use the equation PV = work, where P is the pressure, V is the volume, and work is the work done on the gas. In this case, the gas is losing heat.
First, let's find the initial volume of the gas. We are given the area of the piston, which is 3.06 * 10^-2 m^2. Assuming the piston drops vertically, the initial volume is given by V = area * distance.
Now, let's find the final volume of the gas. Since the gas is losing heat, it will contract. We can assume the gas behaves ideally, so the final volume can be found using the ideal gas law:
PV = nRT
Since we are given the pressure outside the cylinder (1.02 * 10^5 Pa), we can use this pressure as the final pressure (P) in the equation. The temperature (T) and the amount of gas (n) remain constant. Thus, the initial and final volumes are related by:
V1 = V2 * (P2 / P1)
Solving for V2, we get:
V2 = V1 * (P1 / P2)
Now, we know the initial and final volumes of the gas. The distance through which the piston drops (d) can be found by subtracting the final volume from the initial volume:
d = V1 - V2
Substituting the values of V1 and V2, we can calculate the distance through which the piston drops.