2 moles of an ideal gas at 300K and 5 bar is expanded adiabatically at a constant pressure of 1 bar till the volume doubles. Cv = 3R/2. Calculate w, q, dE, dH and change in T.(adsbygoogle = window.adsbygoogle || []).push({});

I found that w = dE = -997.78J

In class we solved dH by:

dH = CpdT = -1.66kJ

(we found Cp by Cp=Cv+nR ; we found dT by Cv=dE/dT)

I was wondering if we could solve dH another way:

dH = dE + d(PV)

However, I get a different answer..

dH = dE + Pext dV (because pressure is constant)

dH = -997.78J +-997.78J (because Pext dV is work)

dH = -2.00 kJ

I also tried another way:

dH = dE + d(nRT)

dH = -977J + (2mol)(8.3145J/K)(-80K)

dH = -1.08kJ

I am not getting the same value for each of the different ways I am using.... Am I doing something wrong or am I not allowed to solve dH using the other methods?

Also, at constant pressure, isn't dH = qrev.. and q=0, so shouldn'tdH = 0?

Thank you

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# Solving Enthaply by other ways?

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