Tag: behaviour of perfect gas and kinetic theory of gases

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Questions Related to behaviour of perfect gas and kinetic theory of gases

Multiple choice physics behaviour of perfect gas and kinetic theory of gases degree of freedom: law of equipartition of energy law of equipartition of energy law of equipartition of energy and mean free path

A mixture Of $n _ { 2 }$ moles of mono atomic gas and $n _ { 2 }$ moles of diatomic gas has $\frac { C _ { p } } { C _ { V } } = y = 1.5$

  1. $n _ { 1 } = n _ { 2 }$

  2. $2 n _ { 1 } = n _ { 2 }$

  3. $n _ { 1 } = 2 n _ { 2 }$

  4. $2 n _ { 1 } = 3 n _ { 2 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\begin{array}{l} As\, { y _{ mix } }=\dfrac { { { C _{ { p _{ mix } } } } } }{ { { C _{ { v _{ mix } } } } } } where\, { C _{ { p _{ mix } } } }=\dfrac { { { n _{ 1 } }{ C _{ { p _{ 1 } } } }+{ n _{ 2 } }{ C _{ { p _{ 2 } } } } } }{ { { n _{ 1 } }+{ n _{ 2 } } } }  \ and\, \, { C _{ { v _{ mix } } } }=\dfrac { { { n _{ 1 } }{ C _{ { v _{ 1 } } } }+{ n _{ 2 } }{ C _{ { v _{ 2 } } } } } }{ { { n _{ 1 } }+{ n _{ 2 } } } }  \ So,\, \, { y _{ mix } }=\dfrac { { { n _{ 1 } }{ C _{ { p _{ 1 } } } }+{ n _{ 2 } }{ C _{ { p _{ 2 } } } } } }{ { { n _{ 1 } }+{ n _{ 2 } } } }  \ Given\, ,\, for\, monoatomic\, { C _{ p } },\dfrac { 5 }{ 2 } R\, and\, { C _{ { v _{ 1 } } } }=\dfrac { 3 }{ 2 } R \ For\, diatomic\, { C _{ { p _{ 2 } } } }=\dfrac { { 7R } }{ 2 } \, and\, { C _{ { v _{ 2 } } } }=\dfrac { 5 }{ 2 } R \ { y _{ mix } }=\dfrac { { { n _{ 1 } }\times \dfrac { 5 }{ 2 } R+{ n _{ 2 } }\times \dfrac { 7 }{ 2 } R } }{ { { n _{ 1 } }\times \dfrac { 3 }{ 2 } R+{ n _{ 2 } }\times \dfrac { 5 }{ 2 } R } } =\dfrac { { 5{ n _{ 1 } }+7{ n _{ 2 } } } }{ { 3{ n _{ 1 } }+5{ n _{ 2 } } } } =\dfrac { 3 }{ 2 }  \ 10{ n _{ 1 } }+14{ n _{ 2 } }=9{ n _{ 1 } }+15{ n _{ 2 } } \ { n _{ 1 } }={ n _{ 2 } } \ Hence, \ option\, \, A\, \, is\, correct\, \, answer. \end{array}$

Multiple choice physics behaviour of perfect gas and kinetic theory of gases degree of freedom: law of equipartition of energy law of equipartition of energy law of equipartition of energy and mean free path

Three perfect gases at absolute temperatures $T _1, T _2$ and $T _3$ are mixed. The masses of their molecules are $m _1, m _2$ and $m _3$ and the number of molecules are $n _1, n _2$ and $n _3$ respectively. Assuming no loss of energy, the final tempreture of the mixture is 

  1. $\dfrac{T _1 + T _2 + T _3}{3}$

  2. $\dfrac{n _1T _1 + n _2T _2 + T _3 T _3}{n _1 + n _2 + n _3}$

  3. $\dfrac{n _1T _1^2 + n _2T _2^2 + n _3 T _3^2}{n _1 T _1 + n _2 T _2 + n _3 T _3}$

  4. $\dfrac{n _1^2T _1^2 + n _2^2T _2^2 + n _3^2 T _3^2}{n _1 T _1 + n _2 T _2 + n _3 T _3}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The final temperature of a mixture of non-reacting gases is the weighted average of their temperatures based on the number of moles (or molecules) of each gas, assuming equal degrees of freedom. The formula is T_final = (n1*T1 + n2*T2 + n3*T3) / (n1 + n2 + n3).

Multiple choice physics behaviour of perfect gas and kinetic theory of gases degree of freedom: law of equipartition of energy law of equipartition of energy law of equipartition of energy and mean free path

The law of equipartition of energy is applicable to the system whose constituents are :

  1. in random motion

  2. in orderly motion

  3. at rest

  4. moving with constant speed
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
The original idea of equipartition  is to assign a average kinetic energy to each degree of freedom if we take transnational case only. So if there would be ordered motion instead of random motion then there is no need of equilibrium theorem  as it would be useless to take about averages as they all are in ordered motion.
Hence, the answer is in random motion.
Multiple choice physics behaviour of perfect gas and kinetic theory of gases degree of freedom: law of equipartition of energy law of equipartition of energy law of equipartition of energy and mean free path

The heat capacity at constant volume of a sampleof 192 g of gas in a container of volume 80$\mathrm { L }$ at atemperature of $402 ^ { \circ } \mathrm { C }$ and at a pressure of$4.2 \times 10 ^ { 5 } \mathrm { Pa }$ is 124.5$\mathrm { JK }$ . The number of thedegrees of freedom of the gas molecules is 

  1. 3

  2. 5

  3. 7

  4. 6
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics behaviour of perfect gas and kinetic theory of gases degree of freedom: law of equipartition of energy law of equipartition of energy law of equipartition of energy and mean free path

Statement -1 : The total translational kinetic energy of all the molecules of a given mass of an ideal gas is 1.5 times the product of its pressure and its volume.
and
Statement -2: The molecules of a gas collide with each other and the velocities of the molecules change due to the collision.

  1. Statement - 1 is True, Statement -2 is True, Statement -2 is a correct explanation for Statements-1

  2. Statement - 1 is True, Statement -2 is True, Statement -2 is NOT a correct explanation for Statements-1

  3. Statement - 1 is True, Statement -2 is False

  4. Statement - 1 is False, Statement -2 is True
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice physics behaviour of perfect gas and kinetic theory of gases degree of freedom: law of equipartition of energy law of equipartition of energy law of equipartition of energy and mean free path

The mass of glucose that should be dissolved in 100 g of water in order to produce same lowering of vapour pressure as is produced by dissolving 1 g of urea (mol. Mass = 60) in 50 g of water is : (Assume dilute solution in both cases) 

  1. 1 g

  2. 2 g

  3. 6 g

  4. 12 g
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Relative lowering of vapor pressure is proportional to the mole fraction of the solute. For dilute solutions, (n_glucose / n_water1) = (n_urea / n_water2). (m_glucose / 180) / (100 / 18) = (1 / 60) / (50 / 18). Solving for m_glucose gives 6 g.

Multiple choice physics behaviour of perfect gas and kinetic theory of gases degree of freedom: law of equipartition of energy law of equipartition of energy law of equipartition of energy and mean free path

In a process $PT=Constant$, if molar heat capacity of a gas is $C=37.35J/mol=K$, then find the number of degrees of freedom of molecules in the gas.

  1. $n=10$

  2. $n=5$

  3. $n=6$

  4. $n=7$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a process PT = constant, the molar heat capacity is C = Cv + R / (1 - x), where PV^x = constant. Since PT = constant, P(PV/nR) = constant, so P^2V = constant, or PV^0.5 = constant. Thus x = 0.5. C = (f/2)R + R / (1 - 0.5) = (f/2)R + 2R. With C = 37.35 and R = 8.314, 37.35 = R(f/2 + 2) => 4.49 = f/2 + 2 => f/2 = 2.49 => f = 5.