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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)

User Cusspvz
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2 Answers

3 votes

Answer: -550.059 J

Step-by-step 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* (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* 101.3=-550.059J

The negative sign indicates the system is doing work.

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

User Magnus Sjungare
by
7.5k points
3 votes
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
User Syed Mehtab Hassan
by
8.4k points
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