Tag: thermal physics

Questions Related to thermal physics

An ideal gas is at a temperature  $T$  having molecules each of mass  $m .$  If  $k$  is the Boltzmann's constant and  $2 \mathrm { kT } / \mathrm { m } = 1.40 \times 10 ^ { 5 } \mathrm { m } ^ { 2 } / \mathrm { s } ^ { 2 } .$  Find the percentage of the fraction of molecules whose speed lie in the range  $324\mathrm { m } / \mathrm { s }$  to  $326\mathrm { m } / \mathrm { s } .$

  1. $0.52 \%$

  2. $0.43 \%$

  3. $0.21 \%$

  4. $0.14 \%$


Correct Option: A

In Vander Waal's equation the critical $P _{c}$ is given by

  1. 3b

  2. $\displaystyle\ \frac{a}{27b^{2}}$

  3. $\displaystyle\ \frac{27a}{b^{2}}$

  4. $\displaystyle\ \frac{b^{2}}{a}$


Correct Option: B
Explanation:

The Vander Wall's equation of state is 
$\left(P+ \displaystyle\ \frac{a}{V^{2}}\right)$ $(V-b)$ = $RT$
$P = \displaystyle\ \frac{RT} {V-b} -\displaystyle\ \frac{a}{V^{2}}$
At the critical point, 
$P = P _{C}, V= V _{C}$ and $T= T _{C}$ 
$\therefore P _{C} = \displaystyle\ \frac{RT _{C}}{V _{C}-b} - \frac{a}{V _{C}^{2}}$ .......(i)
At the critical point on the isothermal,
$\displaystyle\ \frac{dP _{C}}{dV _{C}} = 0$
$\therefore 0 = \displaystyle\ \frac{-RT _{C}}{(V _{C}-b)^{2}} + \frac{1a}{V _{C}^{3}}$
$\displaystyle\ \frac{RT _{C}}{(V _{C}-b)^{2}} = \displaystyle\ \frac{2a}{V _{C}^{3}}$ .... (ii)
Also at critical point,
$\displaystyle\ \frac{d^{2}P _{C}}{dV _{C}^{2}} = 0$
$0 = \displaystyle\ \frac{2RT}{(V _{C}-b)^{3}}- \frac{6a}{V _{C}^{4}}$
$\displaystyle\ \frac{2RT _{C}}{(V _{C-b})^{3}}$ = $\displaystyle\ \frac{5a}{V _{C}^{4}}$ ....(iii)
Dividing (ii) by (iii) we get
$\displaystyle\ \frac{1}{2}(V _{C}-b)$ = $\displaystyle\ \frac{1}{3} V _{C}$
$V _{C} = 3b$ ..... (iv)
Putting this value in (ii), we get 
$\displaystyle\ \frac{RT _{C}}{4b^{2}} = \displaystyle\ \frac{2a}{27{b}^{3}}$
$T _{C} = \displaystyle\ \frac{8a}{27bR}$ .......(v)
Putting the value of $V _{C}$ and $T _{C}$ in (i), we get 
$P _{C}$ = $\displaystyle\ \frac{R}{2b}\left(\frac{8a}{27bR}\right)$ = $\displaystyle\ \frac{a}{9b^{2}}$
$= \displaystyle\ \frac{a}{27b^{2}}$

The temperature of an ideal gas at atmospheric pressure is 300K and volume $lm^3$.If temperature and volume become double, then pressure will be

  1. $10^5 N/m^2$

  2. $2\times 10^5 N/m^2$

  3. $0.5\times 10^5 N/m^2$

  4. $4\times 10^5 N/m^2$


Correct Option: A
Explanation:

$\begin{array}{l} \dfrac { { { P _{ 1 } }{ V _{ 1 } } } }{ { { T _{ 1 } } } } =\dfrac { { { P _{ 2 } }{ V _{ 2 } } } }{ { { T _{ 2 } } } }  \ \Rightarrow \dfrac { { { { 10 }^{ 5 } }\times \left( { 1{ m^{ 3 } } } \right)  } }{ { 300K } } =\dfrac { { P\left( 2 \right)  } }{ { 600 } }  \ \Rightarrow P={ 10^{ 5 } }N/{ m^{ 2 } } \ Hence, \ option\, \, A\, \, is\, correct\, \, answer. \end{array}$

Assertion: Real gases do not obey the ideal gas equation.

Reason: In the ideal gas equation, the volume occupied by the molecules as well as the inter molecular forces are ignored.

  1. Both assertion (A) and reason (R) are correct and R gives the correct explanation

  2. Both assertion (A) and reason (R) are correct but R doesnt give the correct explanation

  3. A is true but R is false

  4. A is false but R is true


Correct Option: A
Explanation:

The ideal gas law treats the molecules of a gas as point particles with  perfectly elastic collisions. This works well for dilute gases in many experimental circumstances. But gas molecules are not point masses, and there are circumstances where the properties of the molecules have an experimentally measurable effect.

In the year 1984, the Bhopal gas tragedy was caused by the leakage of

  1. Carbon monoxide

  2. Methyl isocyanate

  3. Nitrogen oxide

  4. Sulphur oxide


Correct Option: B
Explanation:

The Bhopal gas tragedy was an industrial catastrophe that occurred in 1984 at the Methyl isocynate gas ($CH _3NCO$) was leaked from the plant. Union Carbide India Limited (UCIL) pesticide plant in Bhopal, Madhya Pradesh.

A real gas can be approximated to an ideal gas at

  1. Low density

  2. High pressure

  3. High density

  4. Low temperature


Correct Option: A
Explanation:

Real gas can be approximated as ideal gas when the pressure is low and the temperature is high
This means that per unit volume, there are less number of gas molecules because there is less force (pressure) and there is more energy (temperature), so the molecules will tend to move apart
So, density will be low.

If N be the Avogardo's number and R be the gas constant , then Boltzmann constant id given by 

  1. RN

  2. R/N

  3. N/R

  4. I/RN


Correct Option: B

Real gases approaches ideal gas at high temperature and low pressure because

$A$.   Inter atomic separation is large 

$B$.   Size of the molecule is negligible when compared to inter atomic separation 

  1. a & b are true

  2. only a is true

  3. only b is true

  4. a & b are false


Correct Option: A
Explanation:

Generally, a gas behaves more like an ideal gas at higher temperature and lower pressure as the forces against intermolecular forces becomes less significant compared to the particles' kinetic energy, and the size of the molecules becomes less significant compared to the empty space between them.

A sample of an ideal gas occupies a volume V at a pressure P and absolute temperature T, the mass of each molecule is m. The expression for the density of gas is (k= Boltzmann's constant)

  1. $mkT$

  2. $P/kT$

  3. $P/kTV$

  4. $Pm/kT$


Correct Option: D

The equation of state of n moles of a non-ideal gas can be approximated by the equation 
$ (P + \dfrac{an^2}{V^2})(V -nb) = nRT $ 
where a and b are constants characteristics of the gas. Which of the following can represent the equation of a quasistatic adiabat for this gas (Assume that $C _V$ , the molar heat capacity at constant volume, is independent of temperature) ?

  1. $T(V-nb)^{R/C _v}=$ constant

  2. $T(V-nb)^{C _v/R}=$ constant

  3. $ \begin {pmatrix} T + \frac {ab}{V^2R} \end{pmatrix} (V-nb)^{R/C _v} = $ constant

  4. $ \begin {pmatrix} T + \frac {n^2 ab}{V^2R} \end{pmatrix} (V-nb)^{C _v/R} = $ constant


Correct Option: A
Explanation:

For  a reversible adiabatic process, we have $dS = 0$ (Entropy change = 0)


The entropy equation is $TdS = nC _VdT+T(\frac{\partial P}{\partial T}) _VdV$

From the non-ideal gas equation, $(P+\frac{an^2}{V^2})(V-nb)=nRT$
$(\frac{\partial P}{\partial T}) _V=\frac{nR}{V-nb}$

for $dS = 0$, we have
$nC _VdT = -T(\frac{\partial P}{\partial T}) _VdV=-nRT\frac{dV}{V-nb}$
$\Rightarrow \frac{dT}{T} = -\frac{nR}{C _V}\frac{dV}{V-nb}$
$\Rightarrow ln(\frac{T}{T _0})=-\frac{R}{C _V} ln(\frac{V-nb}{V _0-nb})$

$\Rightarrow T(V-nb)^{\frac{R}{C _V}}=T _0(V _0-nb)^{\frac{R}{C _V}}$

i.e., $T(V-nb)^{\frac{R}{C _V}} = \textrm{constant}$