Consider an ideal gas enclosed in a 1.00 L container at an internal pressure of 10.0 atm. Calculate the work, w , if the gas expands against a constant external pressure of 1.00 atm to a final volume of 10.0 L.

The answer to this question is 1,013 Joules or 1.013 kJ.

Explanation:

This process is an isobaric process as pressure is constant throughout.

Work done by gas expansion, W

= P Δ V   ; (where P = pressure, ΔV = change in volume)

W = 10 atm (10 - 1 ) L

W =  10 L. atm ⇒ 10 ×

W = 1,013 Joules ≈ 1. 01 kJ.

when the volume of gas is   changed from   717.73 ml   to 852 ml   the temperature will   change   from 315   c to 452   c

    calculation

this is calculated using the charles law formula

that is,   v1/t1 = v2/t2

          v1=?

        t1 = 315 c   into   kelvin = 315 +273 = 588 k

        v2 = 852   ml

        t2 = 425c + 273 =698 k

make v1   the subject of the formula ( by   multiplying both side by t1)

  v1=( v2 xt1) / t2

v1= (852 ml   x 588 k)/ 698 k= 717.73 ml