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A 10.0 mol sample of an IG with a CV,m = 1.5R undergoes reversible isothermal compression...

A 10.0 mol sample of an IG with a CV,m = 1.5R undergoes reversible isothermal compression from 0.200 to 0.500 atm at 373K.

a. Calculate the initial and final volumes.

b. q

c. w

d. ∆U

Solutions

Expert Solution

ANSWER=   GIVEN DATA = P1=0.200 ATM, P2=0.500 ATM, T=373 KELVIN, n=10 MOLE, R=1.987 (CONSTANT)

NOW WORK IN REVERSIBLE ISOTHERMAL COMPRESSION =2.303nRTlog10 (P2/P1)

1 INITIAL AND FINAL VOLUME ===> ITS ALWAYS BE INVERSE OF PRESSURE ACCORDING TO BOYAL LAW OF GASES BUT WORK SHOULD BE SAME AS FORMULA =2.303nRTlog10(P2/P1)=2.303nRTlog10(V1/V2)

so that V1 =INITIAL VOLUME =0.500 LITER, AND FINAL VOLME v2=0.200 LITER

2..HERE PROCESS IS ISOTHERMAL SO T=0 SO NO CANGE IN INTERNAL ENERGY =U =0

according to first law of thermodynamic U=q - W so that 0=q- W so q=W

so here q=6.987 kcal (see third point work value because both will be same)

3...work ==>

         SO USE GIVEN DATA TO FIND WORK=2.303*10*1.987*373*log10(0.500/0.200)

           AFTER CALCULATION RESULT IS ===>6987 CAL=6.987 KCAL

4. internal energy U=0

       

queen_honey_blossom answered 2 years ago
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