Helium in a toy balloon does of work on its surroundings as it expands with a constant pressure?

Helium in a toy balloon does 3.30 X 10^2 J of work on its surroundings as it expands with a constant pressure of 2.51 X 10^5 Pa in excess of atmospheric pressure. The balloon's final volume is 1.5 X 10^-3 m3. From this information, determine the initial volume of the gas in the balloon.|||The work done by a constant-pressure expansion process is given by:





w = P * delta-V





where P is the pressure and delta-V is the change in volume.





w = P*(V_final - V_initial)





V_initial = V_final - w/P





In this case, P = 2.51*10^5 Pa + 1 atm = 2.51*10^5 Pa + 1.01*10^5 Pa





P = 3.52*10^5 Pa





Plugging this, and the other values given in the question into the equation for V_initial above, we have:





V_initial = 1.5*10^-3 m^3 - (3.30*10^2 J)/(3.52*10^5 Pa)





V_initial = 5.62*10^-4 m^3