Question

One mole of an ideal gas is expanded from a volume of 1.00 liter to a volume of 6.43 liters against a constant external pressure of 1.00 atm. how much work (in joules) is performed on the surroundings? ignore significant figures for this problem. (t = 300 k; 1 l·atm = 101.3 j)

Answer 1

We actually don't need the value for temperature. There is a direct formula relating work, pressure and volume. The relationship is:

W = Pexternal(ΔV)

Substituting the values and incorporating the conversion ratio,

W = (1 atm)(6.43 L - 1 L)(101.3 J/1 L·atm)
W = 651.36 J

Answer 2

Answer: -550.059 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 = 1.00 atm

$V_1$ = initial volume = 1.00 L

$V_2$ = final volume = 6.43 L

Putting values in above equation, we get:

$W=-1.00atm\times (6.43-1.00)L=-5.43L.atm$

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

$1L.atm=101.3J$

So, $-5.43L.atm=-5.43\times 101.3=-550.059J$

The negative sign indicates the system is doing work.

Hence, the work done on the surroundings is -550.059 J