Question

One mole of an ideal gas is contained in a cylinder with a movable piston. The temperature is constant at 77°C. Weights are removed suddenly from the piston to give the following sequence of three pressures.

a. P1 = 5.50 atm (initial state)
b. P2 = 2.43 atm
c. P3 = 1.00 atm (final state)

What is the total work (in joules) in going from the initial to the final state by way of the preceding two steps?

Answer 1

The total work is 4957.45J

Step-by-step explanation:

For an ideal gas, at constant temperature the definition of work (W) is

`W = - P.dV = - n.R.T \int\limits^i_f {(dV)/(V) \\W = - n.R.T. Ln ((V_(f))/(V_(I)))\\W = - n.R.T. Ln ((P_(i))/(P_(f)))`

where P is the pressure, V the volume, n the moles number, T the temperature and R the gas constant.

To solve the problem is necessary to replace the two steps in the equation

Stape 1: n = 1 mol, R = 0.082atm.L/K.mol, T = 77ºC = 350K, Pi = 5.50atm and Pf = 2.43atm.

`W_(1) = - 1molx0.082(atm.L)/(K.mol)x350KxLn ((5.50atm)/(2.43atm)) = 23.44atm.Lx101.325(J)/(atm.L) =2375.44J`

Stape 2: n = 1 mol, R = 0.082atm.L/K.mol, T = 77ºC = 350K, Pi = 2.43atm and Pf = 1.00atm.

`W_(2) = - 1molx0.082(atm.L)/(K.mol)x350KxLn ((2.43atm)/(1.00atm)) = 25.48atm.Lx101.325(J)/(atm.L) =2582.01J`

The total work is the sum of the two steps

`W = W_(1) + W_(2) = 2375.44J + 2582.01J = 4957.45J`