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?
Let L be the original length of the cylinder of air. (You don't need to know it; it will cancel out). For a monatomic gas, Cp = (5/2)R, where R is the molar gas constant and Cp is the specific heat at constant pressure. This is a constant pressure heat transfer process.
The temperature change is
delta T = Q/(n Cp), and n = PV/RT
n is the number of moles. This leads to
delta T = (2/5) T/(PAL)
For the movement delta X of the piston, use the energy equation
Work = P A *delta X = Q - delta U
where delta U = Cv * delta T
and Cv = (3/2)RT (for a monatomic gas).
I get
delta X = (2/5) Q/(PA)
In your case, delta Q is -2109 J
Check my logic
asd
Your logic is correct. Here's the step-by-step explanation to find the distance the piston drops:
1. Calculate the number of moles (n) of the gas using the equation n = PV/RT, where P is the pressure outside the cylinder, V is the initial volume of the gas, R is the molar gas constant, and T is the initial temperature of the gas.
2. Calculate the temperature change (delta T) using the equation delta T = (2/5) T / (P A L), where T is the initial temperature of the gas, P is the pressure outside the cylinder, A is the area of the piston, and L is the original length of the cylinder.
3. Calculate the change in internal energy (delta U) of the gas using the equation delta U = Cv * delta T, where Cv is the specific heat at constant volume and is equal to (3/2)R for a monatomic gas.
4. Calculate the work (W) done by the gas using the equation P A * delta X = Q - delta U, where P is the pressure outside the cylinder, A is the area of the piston, delta X is the distance the piston drops, Q is the heat removed from the gas, and delta U is the change in internal energy.
5. Rearrange the equation to solve for delta X: delta X = (Q - delta U) / (P A)
6. Substitute the given values into the equation: Q = -2109 J (negative sign indicates heat removed), P = 1.02 * 10^5 Pa, and A = 3.06 * 10^-2 m^2.
7. Calculate delta X using the given values: delta X = (2/5) * (-2109 J) / (1.02 * 10^5 Pa * 3.06 * 10^-2 m^2)
8. Calculate the final value for delta X.
By following these steps, you will be able to find the distance the piston drops.