Thermal Properties of Matter Test

Result

Thermal Properties of Matter Test - 58

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Question 1
1 / -0

The quantity $$\dfrac{PV}{kT}$$ represents

A
mass of the gas

B
kinetic energy of the gas

C
number of moles of the gas

D
number of molecules in the gas

Solution

The ideal gas equation is given as:
$$PV=nRT$$

The Boltzmann constant $$k$$ is represented as:
$$k=\dfrac{R}{N}$$

Now, substitute the value of the Boltzmann constant :
$$PV=nkNT$$

$$\therefore \dfrac{PV}{kT}=nN$$

It is the number of molecules of the particular sample. SO, option $$D$$ is correct.
Question 2
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Let $$C_v$$ and $$C_p$$ denote the molar heat capacities of an ideal gas at constant volume and constant pressure respectively.Which of the following is a universal constant?

A
$$ \dfrac{C_p}{C_v} $$

B
$$ C_p C_v $$

C
$$ C_p - C_v $$

D
$$ C_p + C_v $$

Solution
Question 3
1 / -0

The temperature of a body is increased from $$27^oC$$ to $$127^oC$$. The radiation emitted by it increases by a factor of :

A
$$(256/81)$$

B
$$(15/9)$$

C
$$(4/3)$$

D
$$(12/27)$$

Solution
Question 4
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Three rods of the same cross-sectional and made of the same material from the sides of a triangle $$ABC$$ as shown. The points $$A$$ and $$B$$ are maintained at temperature $$T$$ and $$\sqrt 2 T$$ respectively in the steady state. Assuming that only heat conduction takes plane, the temperature at point $$C$$ is:

A
$$\left[\dfrac {2\sqrt 2 +\sqrt 3}{2+\sqrt 3}\right]T$$

B
$$\left[\dfrac {3 }{1+\sqrt 2}\right]T$$

C
$$\left[\dfrac {2}{\sqrt 3}\right]T$$

D
$$\left[\dfrac {\sqrt 5}{2}\right]T$$

Solution
Question 5
1 / -0

Three identical rods of same material are joined to from an equilateral triangle. The temperature of end $$A$$ and $$B$$ is maintained constant as $$\sqrt {3}T$$ and $$T$$. The ratio of $$T_C / T_B$$ will be: (Assuming no loss of heat from surfaces)

A
$$\dfrac {1+\sqrt 3}{2}$$

B
$$\dfrac {1+\sqrt 3}{2}$$

C
$$\dfrac {1+\sqrt 2}{2}$$

D
$$\dfrac {1-\sqrt 2}{2}$$

Solution
Question 6
1 / -0

Three rods of identical cross-sectional area and made from the same material form the sides of an equilateral triangle. The point $$A$$ and $$B$$ are maintained at $$T$$ and $$2T$$ respectively. In steady state temperature of point $$C$$ is $$T_c$$. ( Assuming only heat conduction takes place.) The value of $$T_c$$ is

A
$$T/2$$

B
$$T$$

C
$$\dfrac{2}{3}T$$

D
$$\dfrac{3}{2}T$$

Solution

Let $$H_1$$ heat flow from $$2T$$ to $$T_c$$

and, $$H_2$$ heat flow from $$T_c$$ to $$T$$

In steady state,

$$H_1 =H_2$$

$$\dfrac {(2T-T_c)KA}{l}$$ $$=$$ $$\dfrac{T_c- T}KA{l}$$

$$2T-T_c = T_c- T$$

$$ 3T= 2T_c$$

$$T_c= \dfrac {3 T}{2}$$
Question 7
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Consider two rods of same length and different specific heats ($$s_1$$ and $$s_2$$), conductivities $$K_1$$ and $$K_2$$ and areas of cross section ($$A_1$$ and $$A_2$$) and both having temperature $$T_1$$ and $$T_2$$ at their ends. If the rate of heat loss due to conduction is equal, then

A
$$K_1 A_1$$ = $$K_2 A_2$$

B
$$K_2 A_1$$ = $$K_1 A_2$$

C
$$\dfrac{K_1 A_1}{s_1}$$ = $$\dfrac{K_2 A_2}{s_2}$$

D
$$\dfrac{K_2 A_1}{s_2}$$ = $$\dfrac{K_1 A_2}{s_1}$$

Solution
Question 8
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At temperature TK, the pressure of 4.0 g argon in a bulb is P. the bulb is put in a bath having temperature higher by 50 K than the first one. 0.8 g of argon gas had to be removed to maintained original pressure .The temperature T is equal to :

A
510 K

B
200 K

C
100 K

D
73 K

Solution
Question 9
1 / -0

Heat is added to an ideal gas and the gas expands. In such a process the temperature:

A
must always increase

B
will remain the same if the work done equals the hear added

C
must always decrease

D
will remain the same if change in internal energy equals the heat added

Solution
Question 10
1 / -0

The temperature of end $$A$$ of a rod us maintained at $$0^oC$$. The temperature of end $$B$$ is changing slowly such that the rod may be considered in steady state at all time and is given by $$T_B $$ = $$\alpha T$$ ; where $$\alpha$$ is positive constant and $$t$$ is time. Temperature of point $$C$$, at a distance $$x$$ from end $$A$$, at any time is ____.

A
$$\dfrac{\alpha xt}{L}$$

B
$$\dfrac{\alpha xt^2}{2L}$$

C
$$\dfrac{\alpha x^2t}{2L}$$

D
$$\dfrac{\alpha (L-x)}{L}t$$

Solution

According to temperature gradient,

$$\dfrac{T_A - T_x}{x}$$ = $$\dfrac{T_A - T_B}{L}$$

$$\Rightarrow$$ $$\dfrac{0 - T_x}{x}$$ = $$\dfrac{0 - \alpha t}{L}$$

$$\Rightarrow$$ $$T_x$$ = $$\dfrac{ \alpha xt}{L}$$