A sample consisting of 1.00 mol Ar is expanded isothermally at 0 degrees C from 22.4 to 44.8 L (a) reversibly, (b) against a constant external pressure equal to the final pressure of the gas, and (c) freely (against zero external pressure). For the three processes calculate q, w, delta U, and delta H. Assume ideal behavior.

a) so for this process q = -w = 1573.25J, ( deltaH = 0 )

b) q = -w = - ( - 1135.68J ) = 1135.68J

c) q = 0

Explanation:

Given that;

Moles of Ar gas is 1.00 mol

a)

isothermal i.e constant temperature expansion of ideal gas deltaU = 0

now since deltaU = 0 = q + w

0 = q + w

q = -w

now for isothermal reversible expansion.

w = -nRTin(Vf/Vt)

= (1.00mol) ( 8.314K^-1.mol^-1) (273K) in(44.8L/22.4L)

= - 1573.25J

so for this process q = -w = 1573.25J

for a perfect gas at constant temperature, deltaH = 0

dH = D( U + PV )

so at constant temperature dU = 0

and PV = constant for perfect gas; d(PV) = 0

THEREfore deltaH = 0

b)

For isothermal i.e constant temperature expansion of ideal gas

deltaU = 0

for a perfect gas at constant temp deltaH = 0

final pressure of the gas in this case is calculated using ideal gas equation

p = nRT/V

we know that {1pa=1Jm^-3}

= [((1.00mol) ( 8.314K^-1.mol^-1) (273K))/(44.8L)] * (1000 Lm^-3)

= 5.07 * 10^4 Pa

for expansion against an external pressure

w = - Pext . deltav

= - (5.07 * 10^4 Pa) ( 44.8L - 22.4L) (1m^3 / 1000L)

= - 1135.68J

q = -w = - ( - 1135.68J ) = 1135.68J

c)

For this scenerio,

deltaU = 0 and deltaH = 0

for free expansion Pext. = 0

therefore Pext * deltaV = 0 * ( 44.8L - 22.4L) = 0

since q = -w, for this process, q = 0

c, liquids evaporate, creating a gas