Answer to Question #232199 in Physical Chemistry for chan

Question

(a) Calculate the work involved when 4 moles of an ideal gas are expanded reversibly from 5.00
bar to 1.00 bar at a constant temperature of 298 K. (b) Calculate the work done if instead the gas
in (a) is compressed reversibly from 1.00 bar to 5.00 bar at the same temperature. (c) How do the
work values in (a) and (b) in magnitude and sign?

Answer 1

Solution.

If temperature is constant, than work:

$A = n \times R \times T \times ln(\frac{p1}{p2})$

a)

$A = 4*8.31*298 \times ln(\frac{5}{1}) = 15942.3 \ J$

b)

$A = 4*8.31*298 \times ln(\frac{1}{5}) = -15942.3 \ J$

c)

When expanding, the gas performs positive work. And when compressed, the gas performs negative work. Therefore, in both cases, the modulo operation is the same, but it differs in sign.

Answer:

a)

A = 15942.3 J

b)

A = - 15942.3 J

c)

When expanding, the gas performs positive work. And when compressed, the gas performs negative work. Therefore, in both cases, the modulo operation is the same, but it differs in sign.