QuestionDetails: This question relates to the isothermal compression of 1.00 mol

Question

QuestionDetails: This question relates to the isothermal compression of 1.00 mol ofan ideal gas at a
temperature of 273 K and a volume of 10.0 L. In the initialstate, a pressure
is applied to the ideal gas by a mass, M1, and temperature of heldconstant by an ice/water
bath.
M1 is replaced by a new mass that is three times as heavy, 3M1, andthe gas is compressed
isothermally.

----As this process occurs, what happens to the mass of ice in theice/water bath? Will the
mass of ice [H2O(s)] increase or decrease? By how many grams willthe mass change?
----Calculate the work for this same process done via an infinitenumber of steps, instead
of just one step.
---Calculate the change in the mass of ice for theinfinite-step process.
QuestionDetails:

Explanation / Answer

By replacing M1 with the heavier mass 3M1, you are applying agreater force on the gas (also a greater pressure). So work isbeing done on the gas. Because the system is isothermal, internalenergy = heat + work = U(T) = 0. This implies that heat = -work. heat, being negative, is flowing from thesystem to the ice/water bath. With heat flowing to the bath, icemelts. So the mass of ice should change heat = -work =PV                    Vfinal = Vinitial *(Pinitial/Pfinal) = 10 L*(M1/ 3M1) = 10/3L Pinitial = nRT/V = (1mol *0.0821 L-atm/mol-K * 273 K) / 10 L =2.24133 atm Pfinal = 3*Pinitial = 6.72399 atm heat = PV = 6.72399 atm *( 10/3 L - 10 L)*(1 m3/ 1000L)*(101325 Pa/1 atm) = -4542.05525 J heat = -4542 J recall enthalpy of fusion for water: 333.55 J/g so 4542 J *( 1 g / 333.55 J ) = 13.6171489 = 13.6 gramsice melted b) infinite number of steps => Pext ~Pgas      (reversible) Work = - integral [ P*dV] = -nRT * integral [ dV/V] =-nRT*ln(V2/V1) Work = -1 mol *(8.314 J/mol-K * 273 K) * ln( 10/3 L / 10 L) Work = 2493.54448 J Work = 2.49 x 103 J c) 2.49 e3 J * ( 1 g / 333.55 J ) = 7.47577419 g ~ 7.48 grams

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