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Find the work done from the following graph. 92 = 600 Joule 500 K P(N/m2) 350 K 9, V> |
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Answer» n=3 moles, T=27 o C=300KP 1 =10ATM P 2 =1atmWork DONE in isothermal REVERSIBLE process,W=−2.303nRTlog(P 1 /P 2 ) =−2.303×3×8.314×300log( 110 ) =5744.14JWork done against constant pressure of 1atm=P opp (V f −V i )=−1atm( P 2 NRT − P 1 nRT )=−1×3×8.314×300( 11 − 101 )=6734.34J . |
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