An ideal gas at temperature t0 is slowly compressed at constant pressure of 2 atm from a volume of 10 liters to a volume of 2 liters. then the volume of the gas is held constant while heat is added, raising the gas temperature back to t0. calculate the work done on the gas. 1 atm = 1.0*105 pascals and 1 liter = 0.001 m3.

Answer: 1600 J

Explanation:

1) dа ta:

a) ideal gas: ⇒ pV = nRT and work = ∫ pdV

b) slowly compressed ⇒ constant temperature and not heat exchange

c) pressure: p = 2 atm

d) intitial volume: Vi = 10 liters

e) final volumen: Vf = 2 liters.

f) then the volume of the gas is held constant ⇒ not work in this stage.

g) calculate the work done on the gas: W = ?

2) Equation

W = ∫pdV

3) Solution:

Since p = constant, W = p ∫dV = p ΔV = p (Vf - Vi)

p = 2 atm * 1.0 * 10⁵ Pa / atm = 200.000 Pa

Vi = 10 liter * 0.001 m³. / liter = 0.01 m³

Vf = 2 liter * 0.001 m³ / liter = 0.002 m³

W = 200.000 Pa * (0.002 m³ - 0.01m³) = - 1.600 J.

The negative sign means the work is done over the system.

That is all the work in the system because at the second stage the volume is held constant.