Question

Consider a container with a frictionless piston that contains a given amount of an ideal gas.

Let’s assume that initially the external pressure is 2.20 bar, which is the sum of a 1 bar atmospheric pressure and the pressure created by a very large number of very small pebbles that rest on top of the piston. The initial volume of gas is   0.300 L   and the initial temperature is 25°C.

Now, you will increase the volume of the gas by changing the external pressure slowly in a way that guarantees that the temperature of the system remains constant throughout the process. To do this, imagine you remove the pebbles one by one slowly to increase the volume by an infinitesimal amount. Every time you remove a weight you allow the system to equilibrate. Your cylinder is immersed in a water bath at 25°C, which keeps your gas at the same temperature throughout the whole process.

Calculate the work (wideal) performed during this reversible, isothermal expansion of the ideal gas.

please show the steps on how to determine work ideal

Answer 1

Since during expansion, pressure decreases and volume increases, these two parameters are assigned opposite signs. The work done by the gas in an infinitesimal expansion is given by,

dW = -PdV

The total work 'W' done by the gas in expansion from original volume V₁ to final volume V_2, will be the sum of the series of the terms PdV in which the pressure keeps on decreasing gradually.

   W = -PdV = -RTdV/V ( since P= nRT /V)

= -RTln(V₂/V₁) = -RTln(P₁/P₂)    [for one mole of ideal gas]

Here T= 25 ⁰C = 298.15 K , V₁ = 0.300 L , P₁ = 2.20 bar , R = 8.314 *10^-2 Lbar K^-1 mol^-1

   Since the intial pressure 2.20 bar is due to the sum of a 1 bar atmospheric pressure and the pressure created by a very large number of pebbles .The pressure exerted by the pebbles = 2.20 -1 =1.20 bar.After completion of whole process pebbles would disappear and the final pressure becomes 1 bar. So P₂ = 1 bar

W_ideal = - 8.314 * 10^-2 Lbar K^-1 mol^-1 * 298.15 K * ln ( 2.20 bar /1 bar)

   = - 8.314 * 10^-2 Lbar K^-1 mol^-1 * 298.15 K* 0.7884

   = - 19.54 Lbar mol^-1

   [Since P₁V₁ = P₂V_{2 ,} V₂ = (2.20 bar *0.300 L)/1 bar = 0.66 L, Therefore

   W_ideal = - 8.314 * 10^-2 Lbar K^-1 mol^-1 * 298.15 K* ln (0.66/ 0.300) = - 19.54 Lbar mol^-1 ]