Kinetic Theory Test

Result

Kinetic Theory Test - 64

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Weekly Quiz Competition

Question 1
1 / -0

A quantity of heat 'Q' is suppplied to a monotomic ideal gas which ecpands at constant pressure. The fractionm of heat that goes into workdone by the gas is

A
$$2/5$$

B
$$3/5$$

C
$$2/3$$

D
$$1$$

Solution
Question 2
1 / -0

'n' moles of an ideal gas undergoes a process A $$\rightarrow$$B as shown in figure. The maximum temperature of the gas during the process will be :

A
$$\dfrac{9P_0V_0}{nR}$$

B
$$\dfrac{9P_0V_0}{4nR}$$

C
$$\dfrac{3P_0V_0}{2nR}$$

D
$$\dfrac{9P_0V_0}{2nR}$$

Solution

$$\begin{array}{l} From\, the\, figure \\ Equation\, of\, line\, AB \\ y-{ y_{ 1 } }=\dfrac { { { y_{ 2 } }-{ y_{ 1 } } } }{ { { x_{ 2 } }-{ x_{ 1 } } } } \left( { x-{ x_{ 1 } } } \right) \\ P-{ P_{ 0 } }=\dfrac { { 2{ P_{ 0 } }-{ p_{ 0 } } } }{ { { V_{ 0 } }2{ V_{ 0 } } } } \left( { V-2{ V_{ 0 } } } \right) \\ =\dfrac { { -{ P_{ 0 } } } }{ { { V_{ 0 } } } } \left( { V-2{ V_{ 0 } } } \right) \\ P=\dfrac { { -{ P_{ 0 } } } }{ { { V_{ 0 } } } } V+3{ P_{ 0 } } \\ PV=-\dfrac { { { P_{ 0 } } } }{ { { V_{ 0 } } } } V+3{ P_{ 0 } }V \\ nRT=-\dfrac { { { P_{ 0 } } } }{ { { V_{ 0 } } } } V+3{ P_{ 0 } }V \\ T=\dfrac { 1 }{ { nR } } \left( { -\dfrac { { { P_{ 0 } } } }{ { { V_{ 0 } } } } V+3{ P_{ 0 } }V } \right) \\ \dfrac { { dT } }{ { dV } } =0\, \, \left( { for\, \max imum\, temperature } \right) \\ -\frac { { { P_{ 0 } } } }{ { { V_{ 0 } } } } 2V+3{ P_{ 0 } }=0 \\ -\dfrac { { { P_{ 0 } } } }{ { { V_{ 0 } } } } 2V=-3{ P_{ 0 } } \\ V=\dfrac { 3 }{ 2 } { V_{ 0 } }\, \, \left( { condition\, for\, \max imum\, \, temperature } \right) \\ { T_{ \max } }=\dfrac { 1 }{ { nR } } \left( { \dfrac { { { P_{ 0 } } } }{ { { V_{ 0 } } } } \times \dfrac { 9 }{ 4 } V_{ 0 }^{ 2 }+3{ P_{ 0 } }\times \frac { 3 }{ 2 } { V_{ 0 } } } \right) \\ =\frac { 1 }{ { nR } } \left( { -\dfrac { 9 }{ 4 } { P_{ 0 } }{ V_{ 0 } }+\dfrac { 9 }{ 2 } { P_{ 0 } }{ V_{ 0 } } } \right) \\ =\dfrac { 9 }{ 4 } \dfrac { { { P_{ 0 } }{ V_{ 0 } } } }{ { nR } } \end{array}$$
Hence, the option $$B$$ is the correct answer.
Question 3
1 / -0

One mole of an ideal monoatomic gas is heated at a constant pressure of one atmosphere from $${0}^{o}C$$ to $${100}^{o}C$$. Then the change in the internal energy is

A
$$6.56$$ joules

B
$$8.32\times{10}^{2}$$ joules

C
$$12.48\times{10}^{2}$$ joules

D
$$20.80$$ joules

Solution
Question 4
1 / -0

A circular disc of mass $$m$$ and radius $$r$$ is rolling about its axis with a constant speed $$v$$. Its kinetic energy is

A
$$\cfrac{1}{4}mv^2$$

B
$$\cfrac{1}{2}mv^2$$

C
$$\cfrac{3}{4}mv^2$$

D
$$mv^2$$

Solution
Question 5
1 / -0

A monoatomic ideal gas undergoes a process given by $$2dU+3dW=0$$ then the process is:

A
isobaric

B
adiabatic

C
isothermal

D
none of these

Solution
Question 6
1 / -0

An ideal gas is enclosed in a cylinder at pressure of 2 atm and temperature, 300 K. The mean time between two successive collisions is $$6\times { 10 }^{ -8 }s$$. If the pressure is doubled and temperature is increased to 500 K, the mean time between two successive collisions will be close to:

A
$$4\times { 10 }^{ -8 }s$$

B
$$3\times { 10 }^{ -6 }s$$

C
$$2\times { 10 }^{ -7 }s$$

D
$$0.5\times { 10 }^{ -8 }s$$

Solution
Question 7
1 / -0

What amount of heat must be supplied to 35 g of oxygen at room temperature to raise its temperature by $${ 80 }^{ \circ }C$$ at constant volume (molecular mass of oxygen is 32 and R = 8.3 J $${ moI }^{ -1 }{ K }^{ -1 }$$)

A
1.52 kJ

B
3.23 kJ

C
1.81 kJ

D
1.62 kJ

Solution
Question 8
1 / -0

When work done by a system was 10 J, the increase in the internal energy of the system was 30 J. The heat 'q' supplied to the system was :

A
10 J

B
+20 J

C
40 J

D
-20 J

Solution
Question 9
1 / -0

The kinetic energy of molecular translation per unit volume of a gas at 2 atmosphere pressure will be nearly

A
$$3\times { 10 }^{ 5 }J$$

B
$$3\times { 10 }^{ 6 }J$$

C
$$3\times { 10 }^{ 7 }J$$

D
$$3\times { 10 }^{ 8 }J$$

Solution
Question 10
1 / -0

A fixed mass of gas is taken through a process $$A\rightarrow B\rightarrow C\rightarrow A$$. Here $$A\rightarrow B$$ is isobaric $$B\rightarrow C$$ is adiabatic and $$C\rightarrow A$$ is isothermal

The pressure at $$C$$ is given by $$(\gamma=1.5)$$

A
$$\frac{10^5}{64}\,N/m^2$$

B
$$\frac{10^5}{32}\,N/m^2$$

C
zero

D
$${10}^5\,N/m^2$$

Solution