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)

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

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kobenhavn kobenhavn · Answer 2 · 2019-12-17T13:25:25+01:00

Answer: -550.059 J

Explanation:

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

[tex]W=-P\Delta V=-P(V_2-V_1)[/tex]

W = amount of work done = ?

P = pressure = 1.00 atm

[tex]V_1[/tex] = initial volume = 1.00 L

[tex]V_2[/tex] = final volume = 6.43 L

Putting values in above equation, we get:

[tex]W=-1.00atm\times (6.43-1.00)L=-5.43L.atm[/tex]

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

[tex]1L.atm=101.3J[/tex]

So, [tex]-5.43L.atm=-5.43\times 101.3=-550.059J[/tex]

The negative sign indicates the system is doing work.

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