If 6.00 moles of a monatomic ideal gas at a temperature of 235 L are expanded isothermally from a volume of 1.26 L to a volume of 4.08 L. How would you calculate the work (W) done by the gas?

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Answer 1 · 2016-12-28T16:20:37

You would do it like this:

There is no change in the internal energy of an ideal gas during an isothermal process - the temperature is constant.

The 1st Law of Thermodynamics tells us that all the heat added to the system is used to do work.

At a constant temperature the work integral is:

#sf(W=int_(V_i)^(V_f)P.dV)#

#sf(PV=nRT)#

#:.##sf(P=(nRT)/(V))#

So the integral can now be written:

#sf(W=nRTint_(V_i)^(V_f)(dV)/V)#

Integrating and putting in the limits gives:

#sf(W=nRTln[V_f/V_i])#

Putting in the numbers (I assume you mean 235 K):

#sf(W=6.00xx8.31xx235xxln[4.08/1.26])#

I can use litres instead of #sf(m^3)# as this is a ratio.

#sf(W=6.00xx8.31xx235xx1.175=1.38xx10^4color(white)(x)J)#