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Topic: q, w, delta H and U - isobaric  (Read 10403 times)

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Offline bu2012

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q, w, delta H and U - isobaric
« on: September 20, 2011, 07:10:45 PM »
One mole of methane (assume ideal gas) is initially at 25 degrees C and 1 atm is heated at constant pressure until the volume is doubled. The molar heat capacity (in cal/mol) variation with respect to temperature is:
Cp = 5.34 + 11.5*10-3 T(K)
Calculate q, w, delta H, and delta U.

I know that q = delta H = Cp*delta T ; w = -P*delta V ; delta U = Cv*delta T but I am confused on what T(K) means and how I can find a numerical form of Cp in order to solve these. Thanks!

Offline bu2012

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Re: q, w, delta H and U - isobaric
« Reply #1 on: September 20, 2011, 11:08:36 PM »
I think I understand now that T(K) doesn't really add to anything but I still am unsure of how to solve for q and so on. q = dH = CpdT but I don't know dT? and w = -PdV but I don't know V...I only know that it is doubled in the end, so I don't know how I would put that in the equation. Please help anyone!

Offline bu2012

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Re: q, w, delta H and U - isobaric
« Reply #2 on: September 21, 2011, 11:22:45 PM »
Well I figured out V using PV=nRT but I am now wondering how to find dT in order to solve for q? I only know the initial temperature. Is there another equation I need to incorporate in order to solve for the final temperature to move along my process?

Offline jusy1

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Re: q, w, delta H and U - isobaric
« Reply #3 on: September 23, 2011, 02:36:00 PM »
You know that when the ideal gas volume doubles, the final temperature will be half the initial temperature (pressure and number of moles held constant).

Now try to solve the integral to get enthalpy and try to see how you can get the other quantities.

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