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### Practice Problem

Posted: Tue Feb 13, 2018 3:59 pm
Can someone please walk me through the steps of this problem?

You have a system consisting of 0.60 moles of an ideal gas contained in a 50.0L container at 1.0 atm. You just love chemistry to a fault, so you perform a series of steps to the system. First, you perform an isobaric compression of the container to 20.0L. Then, you pressurize the system to 8.0 atm using an isochoric method. Finally, you perform a reversible, isothermal expansion (now at 1015.5 K) on your system back to a 50.0L volume at 1.0 atm. Now, to apply your knowledge, you must calculate ΔU, q, w, and ΔS of the system specifically over the entire process.

### Re: Practice Problem

Posted: Tue Feb 13, 2018 10:58 pm
1). Delta U is zero because the final reaction was the same as the first reaction, and because internal energy is a state function, only the initial and final state matter.
2). Delta S is also zero because of the same reason and there was no temperature change.
3). q=-w because delta U=0

solve for w
Volume change from 50L--> 20L
w=-Pext*deltaV = -1.00atm*(20-50)L = 30L.atm * 101.32 (to convert to j) = 3039.75J

Pressure change 1.0 atm --> 8.0atm
w=-Pext*deltaV = 0 because isochoric means volume is constant

Volume change from 20L-->50L
w=-nRTln(v2/v1) = -0.6mol*8.314J/K.mol*1115.5Kln(50/20) = -5640.4

add up w = -2600.65 J
q=-w so q= +2600.65