Faiq said:
Initial volume = ##M*0.08314*298/4##= 6.19M liters
Final Volume = ##M*0.08314*298/2##= 12.39M liters
OK. So STATE 1 is:
0.25M moles, 298 K, 6.19M liters
STATE 2 is:
0.25M moles, 298 K, 12.39M liters
Before we calculate ##\Delta S## for the system, let's first apply the first law of thermodynamics to determine the
change in entropy of the surroundings in this irreversible process.
For an expansion process, the equation for the work W done by the system on the surroundings is:
$$W=\int{p_{ext}dV}$$ where ##p_{ext}## is the externally applied pressure. In our irreversible process, at time zero, we suddenly drop the external pressure from 1 bar to 0.5 bars, and then hold the external pressure constant while the gas expands, until the system re-equilibrates both mechanically and thermally. So, in our process, ##p_{ext}## is constant at 0.5 bars. Based on this, what do you get for the work W done by the system on the surroundings (in terms of M, expressed in liter-bars)?
What is the change in internal energy for this "isothermal" change between STATES 1 and 2? Based on the first law, what is the amount of heat Q transferred from the surroundings to the system in our irreversible process?