Page 89 - E-Modul Fisling Berbasis STEM_Neat

KUNCI JAWABAN

KEGIATAN BELAJAR 1 : HUKUM TERMODINAMIKA

1. Diketahui:
−1
R = 0,08314 L bar K −1 mol
T = 234,150 K
P1 = 1 bar
P2 = 0,315 bar
V2 = 2V1
3
Cv= R
2
Ditanya: nilai W, T2, dan ∆U?
Penyelesaian:

=
1
1
−1
0,08314 −1 . 273,150
=
1
1
−1
= 22,71
1
= −
1
= − (2 − ) = −
1 1
1
1
1
1
−1
= (0,315 ( ))(22,71 )
1,01325
= −7,06
−1
−1
= −715,4
Proses adiabatis, Q = 0 → ∆U = W = C ∆T
v
3
−1
−1
∆T = (−715,4 ) ( 8,314 −1 )
2
− = −57,4
1
2
= 273,15 − 57,4 = 215,75
2
∆ = ∆

3
∆ = (−57,4)
2
∆U = −715,4 J
−1

Proses isokhorik, ∆V = 0 sehingga ∆W = P. ∆V = 0
∆Q = ∆U + ∆W
1000 = ∆U + 0
1000 = ∆U
Jadi, perubahan energi dalam gas ialah 1000 kalori = 1000 kalori x 4,186 J = 4.186 J

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