Question

Calculate entropy/spontaneous : A piston with radius of 15.6 cm containing 2.86 L of an ideal gas (Cv,m = 3/2 R) at 1.00 atm, 298.15 K undergoes isothermal compression by the addition of a 5.64 kg weight.   

Answer 1

First let´s calculate the the area of the piston

15.6 cm = 0.156 m = radius

A = = 3.1416 * 0.156 * 0.156 = 0.0764 m²

Let´s calculate the force the mass has = 5.64 * 9.81 = 55.32 N , this is by multiplying the kg by the gravity

now remember how to relate force with pressure

Pressure = Force / Area

Pressure = 55.32 / 0.0764 = 724.08 Pascal

Let´s change this value to atm, 724.08 Pa = 0.00714611399 atm

Final pressure = 1 + 0.00715 = 1.007 atm

moles of gas

PV = nRT, p is pressure, V is volume, R is gas constant and T is temperature

1 atm * 2.86 = n * 0.082*298.15

n = 1*2.86/(0.082*298.15) = 0.11698 moles of gas

for volume 2 we use the relationship

P1V1 = P2V2

1 atm * 2.86 = 1.007 * V2

V2 = 2.86 / 1.007 = 2.84 L,

let´s apply formula

S change = 0.1167*0.082 * ln(2.84/2.86) = -0.00006715

According to calculations this is a non spontaneous process