Isothermal Expansion?

You cannot answer this question without additional information (tell your teacher that he/she writes sloppy questions), and you are incorrect that the work done is negative; if the system increases in volume, the work done *by* the system on the surroundings must be greater than or equal to zero.

The differential work done *by* the system *on* the surroundings in an expansion/compression process is given by:

δw = p_external dV

where p_external is the external pressure (i.e., the pressure of the surroundings). To calculate this integral, you need to know the pressure of the surroundings as a function of the volume of the system.

In the case of zero external pressure (expansion against a vacuum), the work done by the system is exactly zero.

In the special case of a *reversible* isothermal expansion, the external pressure is at all times equal to the internal pressure of the system (i.e., the system and surroundings are always at equilibrium). In that case, p_external = n*R*T/V and:

δw = (n*R*T/V) dV

Because n,R, and T (for an isothermal process) are all costant, we can easily integrate this to get:

w = n*R*T*ln(V_final/V_initial)

Furthermore, we know that at all times, T = p*V/(n*R), and we are given the initial values of p, V, and n, so we can calculate the temperature:

T = (3.5*10^5 Pa)*(0.0012 m^3)/((0.53 mol)*(8.314 Pa*(m^3))/(mol*K)) = 95.32 K

The work done by the system is then:

w = (p_initial)*(V_initial) * ln(V_final/V_initial)

= (3.5*10^5 Pa)*(0.0012 m^3)*ln(2.8/1.2) = 356 N*m = 356 J (or 360 J, with the correct number of significant figures, because the number of moles is only given with 2 significant figures)