Question

4. 0.25 moles of an ideal gas expands from 72 mL to 10 L. For each of the following experimental conditions, calculate U, q, and w, each in Joules.

a. Gas expands in a vacuum under isothermal conditions.

b. Gas expands isothermally in Denver, Colorado, where the constant external pressure is 675 Torr.

c. Gas expands isothermally in a reversible manner at 19 °C.

d. Gas expands adiabatically against a constant pressure of 760 Torr

Answer 1

4.a. it is an isothermal and irreversible free expansion.

as it is isothermal process and U = f(T), so, . it is also free expansion, i.e. external pressure is zero. so,

w= -pdV = 0, from first law of thermodynamics,

dU = w+ q, so, q=0 [ as dU and w are zero].

4.b. it is an isothermal and irreversible expansion against a constant pressure.

as it is isothermal process and U = f(T), so, .

675 torr = 0.88 atm pressure

-w = Pext *(final volume - initial volume) = 0.88 atm * (10 L - 0.072 L) = 8.73664 atm lit = 885.28 joule

or, w = -885.28 joule

dU = w+ q, so, q=0 [ as dU is zero]. q = 885.28 joule

c. isothermal and reversible manner

as it is isothermal process and U = f(T), so,

-w = number of mole * R * T ln (final volume / initial volume) = 0.25 * 8.314 joule/mol/K * 292 K * ln (10/0.072) = 2994.35 joule

or, w = -2994.35 joule.

dU = w+ q, so, q=0 [ as dU is zero]. q = 2994.35 joule.

d. it is adiabatic expansion mean q = 0

so dU = w

for adiabatic process PV^γ = K (Constant) Where γ = C_p/C_v

Work done : - W = ∫PdV.... by doing integration, work done can be calculated.

note: the value of γ is not given. so, the above integration can't be completed.