Cuối kỳ Vật lý 2( Tiếng Anh )

FINAL PHY2

1/


2/

n = 0.1 moles
V = 2.5 𝑑𝑚3 = 2.5 𝑥 10−3 𝑚3
3V = 7.5 𝑑𝑚3 = 7.5 𝑥 10−3 𝑚3
P1 = 105 𝑃𝑎
𝑅=

25
𝐽/(𝑚𝑜𝑙𝑒. 𝐾)
3

𝛾 = 1.5
a) T1 =

𝑃𝑉
𝑛𝑅

=

105 𝑥 (2.5 𝑥 10−3 )
0.1 𝑥 8.314

= 300.7 𝐾

P1V1 = P2V2 (Chỗ này giống isothermal(Đẳng nhiệt))

Nhiệt độ lúc này chưa thay đổi mà chỉ có áp suất)
V(after the expansion) = V + 3V = 10 𝑑𝑚3 = 10 𝑥 10−3 𝑚3
P1V1 = P2V2
105 𝑥 (2.5 𝑥 10−3 ) = 𝑃2 𝑥 (10 𝑥 10−3 )
 P2 = 25000 Pa
b) Adibatic process : 𝑇1. 𝑉2𝛾−1 = 𝑇2. 𝑉3𝛾−1
V2 = (10 𝑥 10−3 ) 𝑚3
V3 = ( 2.5 𝑥 10−3 ) 𝑚3 (Because the piston back to its initial
position)
 300.7 𝑥 (10 𝑥 10−3 )1.5−1 = 𝑇2 𝑥 (2.5 𝑥 10−3 )1.5−1
 T2 = 601.4 K

3/

T = 273 K
R = 8.314
P = 1.0 𝑥 10−2 𝑎𝑡𝑚 = 1013.25 Pa
d = 1.24 𝑥 10−5 𝑔/𝑐𝑚3 = (1.24 x 10^-5) x 10^6 : 10^3 = 0.0124 kg/m^3


Pascal đi với kg/m^3
a) vrms = √

3𝑅𝑇
𝑀

=√

3𝑃
𝑑

= √

3 𝑥 0.082 𝑥 273
𝑀

=√

3 𝑥 1013.25
0.0124

= 495 m/s
b) vrms = √

3𝑅𝑇
𝑀

−→ 𝑀 = 0.028 kg/mol = 28 g/mol

c) M = 28g/mol  N2
4/

n = 1 mole


P(A) =

𝑃0

2

=

105
2

Pa

P(B) = P0 = 105 𝑃𝑎
P(C) = 3P0 = 3 x 105 Pa
V(A) = V(B) = V0 = 1 𝑚3
a) From A  B (Isochoric = V constant)
V(A) = V(B) = 1 𝑚3
From B  C (Isothermal = T constant)
P(B) x V(B) = P(C) x V(C)
 105 x 1 = 3 x 105 x V(C)
 V(C) = 1/3 𝑚3
b) Adibatic process from C to A
 P(C) x 𝑉 (𝐶 )𝛾 = 𝑃(𝐴) 𝑥 𝑉(𝐴)𝛾
1

105

3

2

 (3 x 105 ) 𝑥 ( )𝛾 =
1

1

3

6

 ( )𝛾 =

1

 𝛾 = log 1 = 1.631
3

6

𝑥 1𝛾


5/ At very low temperatures, the molar specific heat Cv of many solids
is approximately Cv = 𝐴𝑇 3 , where A depends of the particular
substance. For aluminum, A = 3.15 𝑥 10−5 𝐽 𝑚𝑜𝑙 −1 𝐾 −4 . Find the
entropy change for 4.0 mol of aluminum when its temperature is raised
from 5.0 K to 10.0 K

The formula for an entropy change: dS =

𝒅𝑸

Equation: dQ = NCv(T + dT – T)

 dQ = 𝑁𝐶𝑣 (𝑇)𝑑𝑇 (2)
(1)(2)  dS =

𝑁𝐶𝑣(𝑇)𝑑𝑇
𝑇

Integrate both sides  ∆𝑆 = 𝑁 ∫
We have : Cv(T) = 𝐴𝑇 3

𝐶𝑣(𝑇)𝑑𝑇
𝑇

𝑻

(J) (1)


 ∆𝑆 = 𝑁 ∫

𝐴𝑇 3 𝑑𝑇
𝑇

2

= 𝑁𝐴 ∫ 𝑇 𝑑𝑇 = 𝑁𝐴

= 4 x (3.15 x 10^-5) x

103 −53
3

∆𝑇 3
3

= NA

𝑇23 −𝑇13
3

= 0.03675 J

Other question 5:

𝑇2 𝑑𝑄

The formula of entropy change : ∆S = ∫𝑇1

𝑇

dQ = mc𝑑𝑇
𝑇2 𝑚(𝑎+𝑏𝑇)𝑑𝑇

∆S = ∫𝑇1

𝑇

𝑇2 𝑚𝑎

= ∫𝑇1

𝑇

+ 𝑚𝑏 = 𝑚𝑎𝑙𝑛(𝑇) + 𝑚𝑏𝑇
= 3 x 770 x ln

dS =

𝑑𝑄
𝑇

(J/K) (1)

dQ = mCvdT (2)


(1)(2)  dS =

𝑚𝑐𝑑𝑇
𝑇

With c = a+bT. Thus :
𝑇2 𝑚(𝑎+𝑏𝑇)𝑑𝑇

Integrate both side : ∆𝑆 = ∫𝑇1

𝑇

𝑇2

= ∫𝑇1 (𝑚𝑎𝑇 −1 + 𝑚𝑏)𝑑𝑇

= maln(T) + mbT
= maln(600/300) + mb(600-300)
=3x