Kinetic Theory Test

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Kinetic Theory Test - 61

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Weekly Quiz Competition

Question 1
1 / -0

The mass of hydrogen molecules is $$3.32 \times 10 ^ { - 24 }$$ gm. If $$10 ^ { 23 } \mathrm { H } _ { 2 }$$ molecules strike $$2$$ sq. cm are per second with velocity of $$10 ^ { 5 }$$ cm/sec at an angle of $$45 ^ { \circ }$$ to the normal to wall, then the exerted pressure will be

A
$$2.35 \mathrm { N } / \mathrm { m } ^ { 2 }$$

B
$$23.5 \mathrm { N } / \mathrm { m } ^ { 2 }$$

C
$$235 \mathrm { N } / \mathrm { m } ^ { 2 }$$

D
$$2350 \mathrm { N } / \mathrm { m } ^ { 2 }$$

Solution

$$\begin{array}{l} \Delta P=\left( { 2mv\cos { { 45 }^{ o } } } \right) \times { 10^{ 23 } } \\ =2\times 3.32\times { 10^{ -24 } }\times { 10^{ -3 } }\times \frac { { 10 } }{ { 100 } } \times \frac { 1 }{ { \sqrt { 2 } } } \times 10 \\ \Rightarrow F=\frac { { 2\times 3.32 } }{ { \sqrt { 2 } } } \times { 10^{ -1 } } \\ \therefore pressure=\frac { { 2\times 3.32\times { { 10 }^{ -1 } } } }{ { \sqrt { 2 } \times 2\times { { 10 }^{ -4 } } } } \\ =\frac { { 2\times 3.32 } }{ { 2\sqrt { 2 } } } \times { 10^{ 3 } } \\ =\frac { { 3.32 } }{ { \sqrt { 2 } } } \times { 10^{ 3 } }\, N/{ m^{ 2 } } \\ \approx 2350\, N/{ m^{ 2 } } \\ Hence, \\ option\, \, D\, \, is\, \, correct\, \, answer. \end{array}$$
Question 2
1 / -0

Two moles of helium gas are taken along the path ABCD (as shown in figure). The work done by the gas is

A
$$2000R(1 + ln \frac{4}{3})$$

B
$$500R(3+ln4)$$

C
$$500R(2+ln\frac{16}{9})$$

D
$$1000R(1+ln\frac{16}{9})$$

Solution
Question 3
1 / -0

The volume of an ideal gas is 1 litre and its pressure is equal to 72 cm of mercury column. The volume of gas is made $$ 900 cm^3 $$ by compressing it isothermally. The stress of the gas will be

A
8 cm (mercury)

B
7 cm (mercury)

C
6 cm (mercury)

D
4 cm (mercury)

Solution
Question 4
1 / -0

What is the mass of :

A
$$0.2$$ mole of oxygen atoms?

B
$$0.5$$ mole of water molecules?

C
Both a and b

D
None of these
Question 5
1 / -0

Two kg of a monoatomic gas is at a pressure of $$4 \times 10^4 N/m^2$$ . The density of the gas is $$8 kg /m^3$$. What is the order of energy of the gas due to its thermal motion ?

A
$$10^3 J$$

B
$$10^5 J$$

C
$$10^6 J$$

D
$$10^4 J$$

Solution

Thermal energy of N molecule
$$= N \left(\dfrac{3}{2} kT \right)$$
$$= \dfrac{N}{N_A} \dfrac{3}{2} RT$$
$$= \dfrac{3}{2} (nRT)$$
$$= \dfrac{3}{2} PV$$
$$= \dfrac{3}{2} P \left(\dfrac{m}{8} \right)$$
$$= \dfrac{3}{2} \times 4 \times 10^4 \times \dfrac{2}{8}$$
$$= 1.5 \times 10^4$$
order will $$10^4$$.
Question 6
1 / -0

An open container is placed in atmosphere. During a time interval, temperature of atmosphere increases and then decreases. The internal energy of gas in container

A
First increases then decreases because internal energy depends upon temperature

B
First decreases then increases

C
Remains constant

D
Is inversely proportional to temperature

Solution
Question 7
1 / -0

For ideal monoatomic gas,the universal gas constant R is n times molar heat capacity at constant pressure $$ C_p $$ Here n is :-

A
0.67

B
1.4

C
0.4

D
1.67

Solution
Question 8
1 / -0

Half mole of an ideal monoatomic gas is heated at constant pressure of 1 atm from $${ 20 }^{ 0 }C$$ to $${ 90 }^{ 0 }C$$ . Work done by gas is close to : ( gas constant R=8.31 j / mol -k )

A
146 j

B
581 j

C
73 j

D
291 j

Solution
Question 9
1 / -0

One mole of an ideal gas expands against a constant external pressure of 1 atm from a volume of $$10d{ m }^{ 3 }$$ to a volume of $$30d{ m }^{ 3 }$$. What would be the work done in joules?

A
$$-2026 J$$

B
$$+2026 J$$

C
$$-1947 J$$

D
$$1648 J$$

Solution

$$\begin{array}{l} Work\, done\, for\, isothermal\, ansion,W=-\int _{ { V_{ 1 } } }^{ { V_{ 2 } } }{ pdV } \\ { { Sin } }ce,\, \rho \, is\, cons\tan t, \\ W=-\int _{ { V_{ 1 } } }^{ { V_{ 2 } } }{ dV } \\ W=-\rho \left( { { V_{ 2 } }-{ V_{ 1 } } } \right) \\ W=-1\, atm\left( { 30-10 } \right) L \\ =-20\, atm\, L \\ Now,\, 0.08206\, L\, atm=8.314J\left( { As\, we\, know } \right) \\ -20\, atm\, L=\dfrac { { 8.314 } }{ { 0.08206 } } \times -20J \\ =-2026.32J \end{array}$$
Hence,
option $$(A)$$ is correct answer.
Question 10
1 / -0

N moles of a monoatomic gas is carried round the reversible rectangular cycle ABCDA as shown in the diagram. The temperature at $$A$$ is $$T _ { 0 }$$ . The thermodynamic efficiency of the cycle is

A
$$ 15 \%$$

B
$$ 50 \%$$

C
$$ 20 \%$$

D
$$ 25 \%$$

Solution

$$\begin{array}{l} At\, \, A\, \, { P_{ 0 } }{ V_{ 0 } }=nR{ T_{ 0 } }\to \left( i \right) \\ At\, \, B\, \, 2{ P_{ 0 } }{ V_{ 0 } }=nR{ T_{ 1 } }\to \left( { ii } \right) \end{array}$$
Efficiency of thermodynamic cycle
$$\eta = 1 - \frac{{{T_B}}}{{{T_A}}}$$
$$By\left( i \right)\,\,and\,\,\left( {ii} \right)$$
$$\dfrac{{{T_B}}}{{{T_A}}} = \dfrac{1}{2}$$
$$\eta = 1 - \frac{1}{2} = \dfrac{1}{2} = 50\,\,\% $$
Option B.