CLASS : XIth SUBJECT : PHYSICS
Date : Solutions DPP No. : 1
Topic :- KINETIC THEORY
1 (d)
2 2 2 2 2
𝑣𝑟𝑚𝑠 = 𝑣1 + 𝑣2 + 𝑣3 + 𝑣4 + 𝑣5 = 4.24
5
2 (a)
Rate of cooling proportional to (𝑇4 - 𝑇40), as per Stefan’s Law.
𝑅′ (900)4 - (300)4
∴ =
𝑅 (600)4 - (300)4
94 - 34 34(34 - 1)
= 4 4= 4 4
6 -3 3 (2 - 1)
80 16
= 15 = 3
16
𝑅′ = 𝑅
3
3 (c)
The temperature rises by the same amount in the two cases and the internal energy of
an ideal gas depends only on it’s temperature
𝑈1 1
Hence 𝑈2 = 1
4 (b)
4
𝐸2
𝐸1
=
𝑇2
𝑇1()
273 + 84 4 357 4
=( ) =( ) = 2.0
273 + 27 300
5 (a)
𝑓
Kinetic energy for 𝑚 𝑔 gas 𝐸 = 2𝑚𝑟𝑇
If only translational degree of freedom is considered
3 3
Then 𝑓 = 3⇒𝐸Trans = 2𝑚𝑟𝑇 = 2𝑚( )𝑇
𝑅
𝑀
3 8.3
= × 20 × × (273 + 47) = 2490𝐽
2 32
6 (c)
The number of moles of the system remains same,
𝑃1𝑉1 𝑃2𝑉2 𝑃(𝑉1 + 𝑉2) 𝑃(𝑉1 + 𝑉2)𝑇1𝑇2
+ = ⇒𝑇 =
𝑅𝑇1 𝑅𝑇2 𝑅𝑇 (𝑃1𝑉1𝑇2 + 𝑃2𝑉2𝑇1)
According to Boyle’s law,
(𝑃1𝑉1 + 𝑃2𝑉2)𝑇1𝑇2
𝑃1𝑉1 + 𝑃2𝑉2 = 𝑃(𝑉1 + 𝑉2) ∴ 𝑇 =
(𝑃1𝑉1𝑇2 + 𝑃2𝑉2𝑇1)
7 (b)
Saturated water vapour do not obey gas laws
8 (c)
3𝑅𝑇
𝑣𝑟𝑚𝑠 = ⇒𝑇 ∝ 𝑀 [ ∵ 𝑣𝑟𝑚𝑠,𝑅→constant]
𝑀
𝑇 𝑂 2 𝑀𝑂 2 𝑇𝑂2 32
⇒ = ⇒ = ⇒𝑇𝑂2 = 312𝐾 = 39℃
𝑇𝑁2 𝑀𝑁2 (273 + 0) 28
9 (c)
Boyle’s and Charle’s law follow kinetic theory of gases
10 (b)
3
𝐹= 𝑘𝑇⇒𝐸 ∝ 𝑇
2
12 (a)
In elastic collision kinetic energy is conserved
13 (c)
From the Mayer’s formula
𝐶𝑝 - 𝐶𝑉 = 𝑅 …(i)
𝐶𝑝
and 𝛾 = 𝐶𝑉
⇒ 𝛾𝐶𝑉 = 𝐶𝑝 …(ii)
Substituting Eq. (ii) in Eq. (i) we get
⇒ 𝛾𝐶𝑉 - 𝐶𝑉 = 𝑅
𝐶𝑉(𝛾 - 1) = 𝑅
𝑅
𝐶𝑉 = 𝛾 ― 1
14 (b)
From Andrews curve
15 (a)
The rms velocity of an ideal gas is
3𝑅𝑇
𝑣𝑟𝑚𝑠 =
𝑀
Where 𝑇 is the absolute temperature and 𝑀 is the molar mass of an ideal gas
Since 𝑀 remains the same
∴ 𝑣𝑟𝑚𝑠 ∝ 𝑇
𝑣′𝑟𝑚𝑠 3𝑇
= 𝑇 =
′
𝑣𝑟𝑚𝑠 𝑇 𝑇
⇒𝑣′𝑟𝑚𝑠 = 3𝑣𝑟𝑚𝑠
16 (c)
At constant temperature; 𝑃𝑉 = constant
𝑚
⇒𝑃 ×
𝐷
( )
= constant
𝑃
⇒𝐷 = constant = 𝐾. [𝐷 = Density]
17 (a)
3𝑝 𝑣1 𝜌 16 4
𝑣𝑟𝑚𝑠 = ⇒ = 2= =
𝜌 𝑣2 𝜌1 1 1
18 (a)
The gases carbon monoxide (CO) and nitrogen (N2) are diatomic, so both have
5
equal kinetic energy 2𝑘𝑇, 𝑖𝑒.𝐸1 = 𝐸2.
19 (a)
From ideal gas equation, we have
𝑝𝑉 = 𝑛𝑅𝑇
𝑝𝑉
∴ 𝑛=
𝑅𝑇
Given, 𝑝 = 22.4 atm pressure
= 22.4 × 1.01 × 105 Nm-2,
𝑉 = 2L = 2 × 10-3 m3,
𝑅 = 8.31 J mol-1 - K-1,
𝑇 = 273 K
22.4 × 1.01 × 105 × 2 × 10-3
∴ 𝑛=
8.31 × 273
𝑛 = 1.99 ≈ 2
Mass
Since, 𝑛= Atomic weight
We have,
mass = 𝑛 × atomic weight = 2 × 14 = 28 g
20 (d)
3
Average kinetic energy 𝐸 = 2𝑘𝑇
⇒ 𝐸∝𝑇
Thus, average kinetic energy of a gas molecule is directly proportional to the
absolute temperature of gas.
ANSWER-KEY
Q. 1 2 3 4 5 6 7 8 9 10
A. D A C B A C B C C B
Q. 11 12 13 14 15 16 17 18 19 20
A. C A C B A C A A A D
