During the time 0.325 mol of an ideal gas undergoes an isothermal compression at 22.0∘c, 352 j of work is done on it by the surroundings. part a if the final pressure is 1.76 atm, what was the initial pressure?

Respuesta :

According to the equation of the work done W:
When W = nRT㏑(Pi/Pf) and we have:

W = 352 joule & n = 0.325 & T by kelvin = 22+273 = 295 K & Pf= 1.76

by the substitute: 

352 = (0.325 x 8.314 x 295)㏑(Pi/Pf)
∴㏑(Pi/Pf) = -352 / (0.325 x 8.314 x 295) 
㏑(Pi/Pf) = - 0.44
Pi/Pf = 0.644 
∴ Pi =1.76 x 0.644 = 1.13 atm