[7.0分]4.答案解:p2 = nRT / V2
-3
= [2×8.314×300 / ( 49.2×10)] Pa = 101.39 kPa W = -p2?V = [-101 390×( 49.2-4.92 )×10-3]J = -4 489.55 J 可逆压缩作最小功 Wr = -nRTln( V2 / V1 )
= [-2×8.314×300×ln( 4.92 / 49.2 )] J
= 11 486.2 J
[12.0分]5.答案解:(1), Q1 = 0 (视为绝热) ?U1 = Q1 + W1 = 0
?H1 = ?U1 + ?( pV )= p2V2-p1V1 = 1dm3( 50 ? 106-100 ? 103 )Pa = 49.9 kJ (2)等容变化过程,W2 = 0。 Q2 = ?U2 =
= -123.4 kJ
?H2 = ?U2 + ?( pV )= ?U2 +V( p3 - p2 )
=-123.4 kJ + 1dm3 ( 10 ? 106 Pa-50 ? 106 Pa ) =-163.4 kJ 整个过程的W = 0, Q = Q1 + Q2 =-123.4 kJ Q = ?U1 + ?U2 =-123.4 kJ,
?H = ?H1 + ?H2 =-113.5 kJ [8.0分]6.答案解:物质的量一定,对于可逆绝热过程有:
= 21.830 Cp,m = CV,m+R
= ( 21.830+8.314 ) = 30.144
[15.0分]7.答案解:( 1 )对于理想气体的恒温可逆过程:?U = 0,?H = 0; T1 = p1V1 / R = (10.00×105×0.002 / 8.314) K = 240.6 K
=
= 1×8.314×240.6×ln0.5 J = -1386.6 J
Q = -W = 1386.6 J
( 2 ) Qr = 0, T1p1(1?)/? = T2 p2(1?)/?,? = 1.4
T2 = 197 K
-
-
?U = n ?H = n
(?T) = 5 / 2×8.314×( 197-240.6 ) J = -157.966 J (?T) = 7 / 2×8.314×( 197-240.6 ) J = -221.15 J
W = ?U = -157.966 J
[8.0分]8.答案解: W = -p(外) ( Vg-Vl ) ?-nRT =-
-3
×8.314×( 273.15+80.2 )×10 kJ =-3.76 kJ
Q = 395.0×100×10-3 kJ =39.5 kJ
?H = Q = 39.50 kJ
?U = Q+W = [ 39.5 +(-3.76 )] kJ = 35.74 kJ 六、证明(1小题,共12.0分)
[12.0分]1.答案解:(1)范德华方程为: 而 故 (2)因
?H = ?U + ?( pV ) 由范德华方程得:
,积分得:
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