Thermodynamics Test

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Thermodynamics Test - 61

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Question 1
1 / -0

Pressure versus temperature graphs of an ideal gas are as shown in figure. Choose the wrong statement:-

A
Density of gas is increasing in graph (i)

B
Density of gas is decreasing in graph (ii)

C
Density of gas is constant in graph (iii)

D
None of these
Question 2
1 / -0

If the temperature difference $$T_1 - T_2$$ is $$120^oC$$. the temperature difference between points A & B is

A
30

B
45

C
60

D
75

Solution
Question 3
1 / -0

When volume of system is increased two times and temperature is decreased half of its initial temperature, then pressure becomes

A
$$2\ times$$

B
$$4\ times$$

C
$$1/4\ times$$

D
$$1/2\ times$$

Solution
Question 4
1 / -0

A rigid diatomic ideal gas undergoes an adiabatic process at room temperature,. The relation between temperature and volume of this process is $$TV^x$$ = constant, then $$x$$ is :

A
$$\dfrac{5}{3}$$

B
$$\dfrac{2}{5}$$

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

D
$$\dfrac{3}{5}$$

Solution

For adiabatic process :$$TV^{\gamma-1}=$$ constant
For diatomic process : $$\gamma -1=\dfrac{7}{5}-1$$
$$\therefore x=\dfrac{2}{5}$$
Question 5
1 / -0

An ideal gas of adiabatic exponent $$\gamma $$ is supplied heat at a constant pressure. Then ratio dQ : dQ : dU is equal to :

A
$$\gamma :\gamma -1:\dfrac { 1 }{ \gamma } $$

B
$$1\quad :\quad 1:\quad \gamma \quad :\quad 1$$

C
$$\gamma \quad :\quad \gamma -1\quad :\quad 1$$

D
$$\gamma \quad :\quad 1:\quad \gamma \quad -\quad 1$$

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

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

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

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

Solution
Question 7
1 / -0

A vessel contains $$N$$ molecules of a gas at temperature $$T$$. Now the number of molecules is doubled, keeping the total energy in the vessel constant. The temperature of the gas is.

A
$$T$$

B
$$2T$$

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

D
$$\sqrt 2 T$$

Solution

Total energy$$=nkT$$
$$n'=2n$$ (Since the 'P' is in closed box)
$$T'=\cfrac { T }{ 2 } $$
$$PV=nRT$$
$$\therefore T'=\cfrac { T }{ 2 } $$
Question 8
1 / -0

At ordinary temperature, the molecule of an ideal gas has only translational and rotational kinetic energies. At high temperatures, they may also have vibrational energy. As a result of this at higher temperatures ( = molar heat capacity at constant volume)

A
$${ C }_{ v }=\cfrac { 3 }{ 2 }$$ R for a monoatomic gas

B
$${ C }_{ v }>\cfrac { 3 }{ 2 }$$ R for a monoatomic gas

C
$${ C }_{ v }<\cfrac { 5 }{ 2 }$$ R for a diatomic gas

D
$${ C }_{ v }=\cfrac { 5 }{ 2 }$$ R for a diatomic gas

Solution

Vibrational Kinetic energy of a monoatomic gas = 0 at all temperatures. So,
$$Cv = \dfrac{3}{2}R,\,\,So\,\,Cv = \dfrac{3}{2}R$$
for a monoatomic gas even at high temperature also. In case of a diatomic gas
$$Cv = \dfrac{3}{2}R$$ at low temperatures while $$Cv > \dfrac{3}{2}R$$ at high
temperatures due to vibrational Kinetic energy.
Question 9
1 / -0

If the number of molecules of $$\mathrm { H } _ { 2 }$$ are double than that of $$\mathrm { O } _ { 2}, $$ then ratio of mean kinetic energy per molecule of hydrogen to that of oxygen at 300$$\mathrm { K }$$ is

A
1 : 1

B
1 : 2

C
2 : 1

D
1 : 16

Solution
Question 10
1 / -0

The equation of process in terms of volume and temperature is:

A
$$\cfrac V T={constant}$$

B
$$\cfrac V {\sqrt T}={constant}$$

C
$$V T={constant}$$

D
$$ V \sqrt T={constant}$$

Solution