Calculate the work, w, (in j) when 0.6 litre of an ideal gas at an initial pressure of 91.5 atm is expanded isothermally to a final pressure of 2.15 atm against a constant external pressure of 2.15 atm.
i got -130.71 j, but apparently it is wrong. is greatly appreciated.

To calculate work, in thermodynamics, the equation would be as follows:

W = - integral of PdV

P is a function of volume so we cannot really take the P out of the integral. INstead, we can use knowledge on ideal gas to relate P and V.

From the ideal gas equation,

P = nRT / V = k/V

91.5 = k / 0.6
k = 54.9

P = 54.9 / V

W = - integral of ( 54.9/V) dV

Integrating the expression from 91.5 to 2.15, we obtain

W = 205.92 Joules

U whay do you need with?