9. Jawaban: c ΔU = Q – – W a.
ΔU = +Q – (+W ) = 200 J – 200 J = 0
b.
ΔU = –Q – (–W ) = –800 J – (–100 J) = –700 J
c.
ΔU = –Q – (–W ) = –200 J – (–800 J) = 600 J
d.
ΔU = –Q – (+W ) = –400 J – (+200 J) = –600 J
e.
ΔU = +Q – (–W ) = +400 J – (–100 J) = 500 J
10. Jawaban: Diketahui:
d
p = 3 × 105 N/m2 V 1 = 2 × 10–3 m3 V 2 = 4 × 10–3 m3
T 1 = 47°C = 320 K Ditanyakan: n Jawab: W = p (V 2 – V 1)
= (3 × 105 N/m2)(4 × 10–3 m3 – 2 × 10–3 m3) = 600 joule ⎛ 4 × 10−3 m3 ⎞ ⎛ V 2 ⎞ ⎟ (T 1) = ⎜⎜ 3 ⎟ −3 ⎟ (320 K) = 640 K ⎝ V 1 ⎠ ⎝ 2 × 10 m ⎠
T 2 = ⎜
W = n R Δ R ΔT W
n = R ΔT
=
≈ 0,23 mol
Jadi, jumlah zat kira-kira 0,23 mol. a
Diketahui:
m = 80 g T 1 = 27°C = 300 K T 2 = 107°C = 380 K Q p = 2,4 × 103 J
Ditanyakan: C p Jawab: C p =
Q p
ΔT
=
2, 4 × 10 3 J (380 − 300) K
=
2, 4 × 103 J 80 K
= 30 J/K Jadi, kapasitas kalor gas nitrogen pada keadaan tersebut sebesar 30 J/K.
d
Diketahui:
m T 1 T 2 C v V p
= 800 g = 0,8 kg = 30°C = 303 K = 90°C = 363 K = 200 J/kg K = 250 J/kg K
Ditanyakan: W Jawab: C = m m c c W = (C p – C v)(T 2 – T 1) = m (c p – c v)( T 2 – T 1) = (0,8 kg)(250 J/kg K – 200 J/kg K)(363 K – 303 K) = (0,8 kg)(50 J/kg K)(60 K) = 2.400 J = 2,4 kJ Jadi, usaha gas sebesar 2,4 kJ. 13. Jawaban: Diketahui:
d
ΔT = 100 K W = 60 kJ Q p = 80 kJ
Ditanyakan: C v Jawab: W = Q p – Q v 60 kJ = 80 kJ – Q v Q v = 20 kJ = 20.000 J C v =
600 joule joule (8,31 (8,314 4 J/mo J/mol. l.K) K)(6 (640K 40K − 320 320 K)
11. Jawaban:
12. Jawaban:
=
Q v ΔT 20.00 20.000 0J 100 100 K
= 200 J/K Jadi, kapasitas kalor gas saat volume tetap 200 J/K. 14. Jawaban: Diketahui:
a
p = 120 kPa = 1,2 × 105 Pa V 1 = 2 L = 2 × 10–3 m3 V 2 = 3,5 L = 3,5 × 10–3 m3 ΔU = 430 joule Ditanyakan: Q Jawab: – p (V 2 – V 1) ΔU = Q –
430 J = Q – 430 – (1,2 × 105 Pa)(3,5 × 10–3 m3 – 2 × 10–3 m3) 430 43 0 J = Q – – 180 J Q = +610 J Jadi, tanda positif menunjukkan bahwa gas menerima kalor sebesar 610 J. 15. Jawaban: Diketahui:
e
Q = +300 kalori = 1.260 J ΔU = +250 kalori = 1.050 J ΔV = 2,5 L = 2,5 × 10–3 m3 Ditanyakan: p
Fisika Kelas XI
7
