[Hockey66]

1.4 moles of an ideal gas with Cv = 3R/2, initially confined to a container of volume 9 litre and at a temperature 20 C, expand (or are compressed) against a constant external pressure of 1.1 atm until the final pressure of the gas is equal to the external pressure and the final temperature of the gas is equal to the temperature of the surroundings. During this process the system does 3,680 J of work on the surroundings. Calculate the change in entropy, Delta S, of the gas (in J / K).

I'm kind of thrown off by the T(final) = T(surroundings).  I calculated V(final) using w=-p(ex)(dV) and used that to calculate T(final) using the ideal gas law.  I get 402.3K which seems weird to me that it would be higher.  Am I supposed to put in the 3/2R instead of just R when dividing to solve for T? In that case I get 268.2K.  After getting the T(final) can I just plug all the data into deltaS = nCvln(Tf/Ti) + nRln(Vf/Vi)?  Thanks, and no rush.. just need some clarification.  Any tips for solving problems of this sort?

[DevaDevil]

I will here assume the work done by the system is not merely the expansion work, but also the heat transfer. Please be sure if this assumption is correct (if work is merely the expansion work, then your approach is correct)

dQ = dU + p_e dV
work = change in internal energy (temperature dependent) + p dV
dQ = n C_v dT + p_e dV
dQ = 3/2 n R dT + p_e dV
dQ = 3/2 nR (T_e-T_b) + p_e (V_e-V_b)
dQ = 3/2 nR T_e + p_e V_e - ( 3/2 nR T_b + p_e V_b )
dQ = 3/2 nR T_e + p_e V_e - ( 5116 + 1003 )
dQ = 3/2 nR T_e + p_e V_e - ( 6119 )

and: p_e V_e = nR T_e
give:

dQ = 3/2 p_e V_e + p_e V_e - ( 6119 )
dQ = 5/2 p_e V_e - 6119
3680 = 2.79.10⁵ V_e - 6119

V_e = 0.035 m³ = 35 l.
and
T_e = p_e V_e / nR = 337K.

the temperature rize is not unexpected, since the pressure drop of the system (from 3.7 atm to 1.1) is relatively smaller than the volume increase (9 to 35 liter)