Question

1.5 moles of an ideal gas, for which the molar heat capacity Cv.m = 3/2R, initially at 25.0°C and 1 atm undergoes a two-stage transformation. Calculate q. w. Au in kJ for the complete process (a+b). a. The gas is expanded isothermally and reversibly until the volume doubles b. Beginning at the end of the first stage, the temperature is raised to 100.0°C at constant volume. 9 -2.57 ki A. W = 2.57 kJ AU = 5.14 k) = +2.57 kJ B.W = -2.57 kJ AU - 1.40 kJ 9 = + 3.97 kJ C. W = -2.57 kJ AU = 1.40 kJ -- 3.97 kJ DW 2.57 kJ AU -1.40 kJ

Answer 1

Giveni fan n=1.5 moles C, n = 3/2 R = T T = 25.0°c = 298 k P = T = CS Scanned with CamScanner latim 100°c 373k

- 1.5 x 0.082057 x 298. V = nRT 36.68 L 3 36.0 ✓ Also, 2 a 2v = 2x36.68L V₂ = 73.36 L for 0 - ② (lso thermal process) pivi - Po V2 = 0.5 atm Volume Remains constant - ₂ = 73.36 So, la = 0.5 P = 0.62 atry Scanned with CamScanner

(atm) [ 0 (36.4, 1) pe o = 298 K (73.4, 0.5) © (7314,0.5) (L) Ics Scanned with CamScanner

Now, W, = nRT la 12 24 [ for O-) 21.5802083 14 X 298x In a = 2575 J = 2.57kJ W₂ = OKJ [for ③ - ④ As volume is constent wa witwe = 2.57 to Tw= 2.57 kJ Jew, AU & 03 [for 0 - ① ] As temperature is constant AU= nCmAT - n 3 R AT = 1.5x3 x 8.3144 (373-298) 1402.98 5 - 1.40 kJ AUF AU, too = 11.40 kJ

Now as per prot law qe ou + w q = lino t 2.57 = 3.97 kr octo be correct It seems aptin Scanned with CamScanner

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