Question

What will be the work done when one mole of a gas expands isothermally from 15L to 50L against a constant pressure at 1atm at 25 C

Answer 1

Answer: The work done for the given process is -3546.55 J

Explanation:

To calculate the amount of work done for an isothermal process is given by the equation:

$W=-P\Delta V=-P(V_2-V_1)$

W = amount of work done = ?

P = pressure = 1 atm

$V_1$ = initial volume = 15 L

$V_2$ = final volume = 50 L

Putting values in above equation, we get:

$W=-1atm\times (50-15)L=-35L.atm$

To convert this into joules, we use the conversion factor:

$1L.atm=101.33J$

So, $-35L.atm=-35\times 101.3=-3546.55J$

The negative sign indicates the system is doing work.

Hence, the work done for the given process is -3546.55 J

Answer 2

Answer:

W = -713 cal & -2983.01315 J

Explanation:

The formula used:

W = -2.303 nRT log[Vf / Vi]

Given Data:

n = one mole;

Vi = 15L;

Vf = 50L;

P = 1 atm;

T = 25°C = ( 25 + 273 )K = 298K

Solution:

W = -2.303 RT log( Vf / Vi)

= -2.303 x 8.314 J/K mol x 298 x log ( Vf / Vi )

= -5705.84832 x 0.522878745

= -2983.01315 J

= -( 2983.01315 x 0.239) cal

= -712.940143 cal