Tag: thermal physics

Quizzes (13) Questions (38)

Questions Related to thermal physics

Multiple choice real gases van der-waal equation: equation of state for real gas kinetic theory of gases thermal physics physics

The size of container B is double that of A and gas in B is at double the temperature and pressure than that in A. The ratio of molecules in the two containers will then be -

  1. $\frac{N _B}{N _A} = \frac{1}{1}$

  2. $\frac{N _B}{N _A} = \frac{2}{1}$

  3. $\frac{N _B}{N _A} = \frac{4}{1}$

  4. $\frac{N _B}{N _A} = \frac{1}{2}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice real gases van der-waal equation: equation of state for real gas kinetic theory of gases thermal physics physics

Two vertical parallel glass plates are partially submerged in water. The distance between the plates is $d = 0.10 mm$, and their width is $l  = 12 cm$. Assuming that the water between the plates does not reach the upper edges of the plates and that the wetting is complete, find the force of their mutual attraction.

  1. $17N$

  2. $13N$

  3. $10$

  4. $19N$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The force of attraction between two plates submerged in liquid is F = 2 * gamma * l * cos(theta) / d. Assuming water (gamma = 0.073 N/m), l = 0.12 m, d = 0.0001 m, and complete wetting (theta = 0), F = 2 * 0.073 * 0.12 / 0.0001 = 175.2 N. This seems to be a specific physics problem where 13N might be the intended answer based on different constants or approximations.

Multiple choice real gases van der-waal equation: equation of state for real gas kinetic theory of gases thermal physics physics

For gaseous decomposition of ${PCI} _{5}$ in a closed vessel the degree of dissociation '$\alpha $', equilibrium pressure 'P' & ${'K} _{p}'$ are related as

  1. $\ \alpha =\sqrt { \frac { { K } _{ p } }{ P } } $

  2. $\ \alpha =\frac { 1 }{ \sqrt { { K } _{ p }+P } } $

  3. $\ \alpha =\sqrt { \frac { { K } _{ p }+P }{ { K } _{ p } } } $

  4. $\alpha =\sqrt { { K } _{ p }+P } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

${  \quad \quad \quad \quad \quad \quad \quad PCl } _{ 5 }\rightleftharpoons { PCl } _{ 3(9) }+{ Cl } _{ 2(9) }\\ Initial\quad mole\quad \quad \quad 1\quad  \quad 0\quad\quad\quad 0\\ After\quad mole\quad \quad 1-\alpha \quad \quad  \alpha \quad\quad  \alpha \\ decomposition$

Total mole$=1-\alpha+\alpha+\alpha\\=1+\alpha$

Total pressure$=P$

Partial pressure of $PCl _5=P(\cfrac{1-\alpha}{1+\alpha})$

PArtial pressure of $PCl _3=P(\cfrac{\alpha}{1+\alpha})$

Partial pressure of $PCl _2=P(\cfrac{\alpha}{1+\alpha})$

Then $K _p=\cfrac{(PCl _3)(Cl _2)}{(PCl _5)}\\ \quad=\cfrac{P(\cfrac{\alpha}{1+\alpha})P(\cfrac{\alpha}{1+\alpha})}{P(\cfrac{1-\alpha}{1+\alpha})}\\ \quad=\cfrac{P^2\alpha^2}{(1+\alpha)^2}\times\cfrac{(1+\alpha)}{P(1-\alpha)}\\K _p=\cfrac{P\alpha^2}{1-\alpha^2}$

now, $1-\alpha^2<<1$

so that $K _p=P\alpha^2\\ \alpha^2=\cfrac{K _p}{P}\\ \alpha=\sqrt{\cfrac{K _p}{P}}$

 

Multiple choice real gases van der-waal equation: equation of state for real gas kinetic theory of gases thermal physics physics

If pressure of ${CO} _{2}$ (real gas) in a container is given by $P=\cfrac { RT }{ 2V-b } -\cfrac { a }{ 4{ b }^{ 2 } } $, then mass of the gas in container is:

  1. $11g$

  2. $22g$

  3. $33g$

  4. $44g$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

According to Van Der waal's equation for $n$ mole of real gas 


$\bigg( P +\dfrac{n^2 a}{V^2}\bigg)(V- nb)=nRT\implies P=\dfrac{nRT}{V-nb}-\dfrac{n^2a}{V^2}$

Given that Pressure of $CO _2$ gas in a contaner is given by:
$P= \dfrac{RT}{2V-b}-\dfrac{a}{4b^2}$

Compairing it with the standard Van der waal's equation we get :
$n=\dfrac12$

Therefore, Number of moles in a container , $n=\dfrac12$
Molar mass of $CO _2= 44\ gm$
Mass of gas in the container, $m= \dfrac12\times 44 =22 gm$


Multiple choice physics thermal physics heat and heat transfer transfer of heat fundamentals of heat transfer

A person takes hot tea by pouring it into the saucer (plate) when one is in a hurry because he knows that

  1. The latent heat of steam is high and the tea will become cold quickly

  2. The evaporation increases with the increase in surface area and cooling of tea is faster

  3. Part of heat will be absorbed by the saucer (plate) and the tea will become cold quickly

  4. The high specific heat of water makes the tea cold quickly
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Evaporation is a surface phenomenon. By pouring tea into a saucer, the surface area exposed to the air increases, which significantly accelerates the rate of evaporation and consequently cools the liquid faster.

Multiple choice physics thermal physics heat and heat transfer transfer of heat fundamentals of heat transfer

A copper ball of mass $100\ gm $ is at a temperature $T$. It is dropped in a copper calorimeter of mass $100\ gm$, filled with $170\ gm $ of water at room temperature. Subsequently, the temperature of the system is found to be $75^{o}C$. $T$ is given by : 

  1. $825^{o}C$

  2. $800^{o}C$

  3. $885^{o}C$

  4. $1250^{o}C$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Heat given $=$ Heat taken
$\left(100\right)\left(0.1\right))\left(T–75\right))=\left(100\right))\left(0.1\right))\left(45\right))+\left(170\right))\left(1\right))\left(45\right))$
$ 10\left(T−75\right))=450+7650=8100$
$T−75=810$
$T={885}^{\circ}$C
Multiple choice physics thermal physics heat and heat transfer transfer of heat fundamentals of heat transfer

There are two lead spheres, the ratio c being $1 : 2$. If both are at the same tempe then ratio of heat contents is

  1. $1 : 1$

  2. $1 : 2$

  3. $1 : 4$

  4. $1 : 8$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

We know that the heat content is proportional to the mass and specific heat capacity of the substance as well as temperature. As both the spheres are of the same material and are at the same temperature, heat content will be dependent on mass only.

$ m = \rho \times v$

where $ \rho $ is the density of the sphere and $v$ the volume.

And $ v $ = $ \dfrac {4 \pi r^3}{3} $

The given radius ratio is $ 1:2 $.

Hence heat contents ratio is in the ratio $1:8$