工程竣工决算审计内容丶方法与技巧PPT培训课件

。Р解:( 1 )由式,对定压过程,则? W = -psu ( V2-V1 ) = -psunR ? 因为:p1 = p2 = psu ? 所以:W = -nR ( T2-T1 )= -20.78 kJР1-1-4. 10 mol的理想气体,压力1013 kPa,温度300 K,分别求出定温时下列过程的功:? ( 1 )向真空中膨胀;? ( 2 )在外压力101.3 kPa下体积胀大1 dm3;? ( 3 )在外压力101.3 kPa下膨胀到该气体压力也是101.3 kPa;? ( 4 )定温可逆膨胀至气体的压力为101.3 kPa。Р( 4 ) W = nRTln( p2 / p1 )? = [ 10×8.314×300ln( 101.3 / 1013 ) ] J ? = -57.43 kJР解:( 1 ) W = 0 ( 因psu = 0 )Р( 2 ) W = -101.3×1 J = -101.3 JР( 3 ) W = -psu[ nRT( 1 / p2-1 / p1 ) ]? = [-101.3×10×8.314×300( 1 / 101.3-1 / 1013 )] J ? = 22.45 kJР1-1-5. 10 mol理想气体由25℃,1.0 MPa膨胀到25℃,0.1 MPa,设过程为:( 1 )自由膨胀;( 2 )对抗恒外压力0.1 MPa膨胀;( 3 )定温可逆膨胀。试计算三种膨胀过程中系统对环境作的功。Р解:( 1 ) W = 0Р( 3 ) W = -nRTln( V2 / V1 ) = nRT lnp2 / p1 = -57.05 kJР( 2 ) W = -psuV = -psunRT ? = -0.1 MPa×10 mol×8.314 J·mol1·K1×298 K? ×(1/0.1MPa-1/1.0MPa) ? = -22.30 kJ