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?
To determine the distance the piston drops, we can use the first law of thermodynamics, which states that the change in internal energy of a system is equal to the heat added to the system minus the work done by the system.
In this case, the heat is removed from the gas, so it can be considered negative. The work done by the system is given by the equation: Work = Force × Distance.
The force exerted by the gas can be determined using the ideal gas law, which states that pressure times volume is equal to the number of moles times the gas constant times the temperature. In this case, since the gas is monatomic, the number of moles is equal to the mass of the gas divided by the molar mass.
The volume of the gas is equal to the area of the piston times the distance it moves. Substituting these values into the ideal gas law equation, we can solve for the force exerted by the gas.
Next, we substitute the force into the work equation to determine the work done by the system.
Finally, we can solve for the distance by rearranging the work equation to isolate the distance variable.
Let's plug in the given values and calculate the answer:
Given:
Area of the piston (A) = 3.06 × 10^(-2) m^2
Pressure outside the cylinder (P) = 1.02 × 10^5 Pa
Heat removed from the gas (Q) = -2109 J
First, let's calculate the force exerted by the gas:
Volume (V) = A × distance (d)
Number of moles (n) = mass of gas (m) / molar mass
Using the ideal gas law: P × V = n × R × T
Rearranging the ideal gas law equation to isolate the force (F):
F = P × A × d / (R × T)
Substituting the given values:
F = (1.02 × 10^5 Pa) × (3.06 × 10^(-2) m^2) × d / (R × T)
The work done by the system is:
Work = F × d
Now, we substitute the force equation into the work equation:
Work = [(1.02 × 10^5 Pa) × (3.06 × 10^(-2) m^2) × d / (R × T)] × d
Since we are solving for the distance (d), we rearrange the work equation:
d^2 = [Work × (R × T)] / [(1.02 × 10^5 Pa) × (3.06 × 10^(-2) m^2)]
Substituting the given values:
d^2 = [-2109 J × (R × T)] / [(1.02 × 10^5 Pa) × (3.06 × 10^(-2) m^2)]
Finally, calculate the distance (d):
d = sqrt{[-2109 J × (R × T)] / [(1.02 × 10^5 Pa) × (3.06 × 10^(-2) m^2)]}
Note: Make sure to use the appropriate values for the gas constant (R) and temperature (T) in order to obtain accurate results.