Tag: thermal physics

Questions Related to thermal physics

As per Langmuir model of adsorption of a gas on a solid surface.

  1. The mass of gas striking a surface area is independent of the pressure of the gas

  2. The adsorption can be multilayer.

  3. The rate of desorption does not depend on the pressure.

  4. The rate of desorption does not depend on the surface are adsorbed.


Correct Option: C
Explanation:

In Langmuir's model of adsorption of a gas on solids surfaces. The adsorption at a single site on the surface may invoice multiple molecules at the ame time. The mass of gas striking at a given area of surface is independent of the pressure of the gas.

Under which of the following conditions is the law $pV=RT$ obeyed most closely by a real gas?

  1. High pressure and high temperature.

  2. Low pressure and low temperature.

  3. High pressure and low temperature.

  4. Low pressure and high temperature.


Correct Option: D
Explanation:

At low pressure and high temperature real gas obey PV=RT  i.e. they behave as ideal gas because at high temperature we can assume that there is no force of attraction or repulsion works among the molecules and the volume occupied by the molecules is negligible in comparison to the volume occupied by the gas

1 mole of $SO _2$ occupies a volume of $350 ml$ at $300K$ and $50 atm $ pressure. Calculate the compressibility factor of the gas.

  1. $0.888$

  2. $0.711$

  3. $0.520$

  4. $0.987$


Correct Option: B
Explanation:

Given $P=50 \ atm , \ \ V=350 ml =0.350 \ \ \ litre, \ \ n=1 \ \ mole $ and $ T=300 K$

Now, Compressibility factor $Z= \dfrac{PV}{nRT}$
$\therefore  \ Z= \dfrac{50 \times 0.350}{1 \times 0.082 \times 300}= 0.711$

A real gas behaves like an ideal gas if its.

  1. Both pressure and temperature are high

  2. Both pressure and temperature are low

  3. Pressure is high and temperature is low

  4. Pressure is low and temperature is high


Correct Option: D
Explanation:

At high temperature and low pressure the real gas behaves as an ideal gas.

The behaviour of the gases, which can be easily liquified, is like that of the

  1. triatomic gases

  2. ideal gases

  3. van der Waals gases

  4. all of the above


Correct Option: C
Explanation:

Van der Walls equation takes into account inter atomic forces between gas particles which is not considered in the ideal gas model. Since simplicity of  liquification of a gas depends upon forces between its particles, Van der Walls equation is followed by easily liquifieable gases. 


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

Parameter a takes into account interatomic forces.

The rms speed of the molecules of enclosed gas is V. What will be the ems speed if pressure is doubled, keeping the temperature same ?

  1. 3 V

  2. 4 V

  3. V

  4. 2 V


Correct Option: B

If 2g of helium is enclosed in a vessel at NTP, how much heat should be added to it to double the pressure ? (Specific heat of helium = 3 J/gm K)

  1. 1638 J

  2. 1019 J

  3. 1568 J

  4. 836 J


Correct Option: A
Explanation:

No. of moles, $n=\dfrac{m}{M}=\dfrac{2}{4}=0.5 mol$

The specific heat, $C _V=3J/g.mol K$
$C _V=12 J/mol. K$
At constant volume,
$\dfrac{T _2}{T _1}=\dfrac{P _1}{P _2}$
$T _2=2T _1$
$\Delta T=2T _1-T _1=T _1=273K$
The heat required, $\Delta Q=nC _V \Delta T$
$\Delta Q=0.5\times 12\times 273$
$\Delta Q=1638 J$
The correct option is A.

The diameter of oxygen molecules is $2.94 \times 10^{-10}m $. The Van der Waals gas constant in m$^3$/mol will be

  1. $3.2$

  2. $32$

  3. $32\times 10^{-6}$

  4. $32 \times 10^{-3}$


Correct Option: C
Explanation:

$\displaystyle b = 4N \times \frac{4}{3} \pi \frac{d^3}{8}$ (standard definition of boyles constant)

$\displaystyle = \frac{4 \times 6.02 \times 10^{23} \times 3.14 \times 2.94^3 \times 10^{-30}}{3 \times 8}$

$\displaystyle = 32 \times 10^{-6}$

Read the given statements and choose which is/are on the basis of kinetic theory of gases.

  1. Energy of one molecule at absolute temperature is zero.

  2. $rms$ speeds of different gases are same at same temperature

  3. For one gram of all ideal gases, kinetic energy is same at same temperature.

  4. For one mole of all ideal gases, mean kinetic energy is same at same temperature.


Correct Option: A

Work done by a system under isothermal change from a volume $V _1$ to $V _2$ for a gas, which obeys vander Waals equation $(V - \beta n) \displaystyle \left ( P + \dfrac{an^2}{V} \right ) = n RT$ is

  1. $\displaystyle n RT log _e \left ( \dfrac{V _2 - n \beta}{V _1 - n \beta} \right ) + an^2 \left ( \dfrac{V _1 - V _2}{V _1 V _2} \right )$

  2. $\displaystyle n RT log _{10} \left ( \dfrac{V _2 - \alpha \beta}{V _1 - \alpha \beta} \right ) + \alpha n^2 \left ( \dfrac{V _1 - V _2}{V _1 V _2} \right )$

  3. $\displaystyle n RT log _e \left ( \dfrac{V _2 - n \alpha}{V _1 - n \alpha} \right ) + \beta n^2 \left ( \dfrac{V _1 - V _2}{V _1 V _2} \right )$

  4. $\displaystyle n RT log _e \left ( \dfrac{V _2 - n \beta}{V _1 - n \beta} \right ) + \alpha^2 \left ( \dfrac{V _1 V _2}{V _1- V _2} \right )$


Correct Option: A
Explanation:
Given Vander Waals equation $(V - \beta n)$ ($P$ $+$ $\dfrac {a{n}^{2}} {{V}^{2}}$) $=$ $nRT$
Work done by the system($W$) = $-$ $\int _{{V} _{1}}^{{V} _{2}} {PdV}$
                                           $=$ $\int _{{V} _{2}}^{{V} _{1}} {PdV}$
From the Vander Waals equation:
$P$ $=$ $\dfrac {nRT} {(V - \beta n)}$ $-$ $\dfrac {a{n}^2} {{V}^{2}}$
Substituting $'P'$ in the Work done, we get
$W$ $=$ $\int _{{V} _{2}}^{{V} _{1}}$ ($\dfrac {nRT} {(V - \beta n)}$ $-$ $\dfrac {a{n}^2} {{V}^{2}}$) $dV$
By integrating we get,
$W$ $=$ $[$ $nRT$ $\log _{e}{(V - \beta n)}$ $-$ ($(-)\dfrac {a{n}^{2}} {V}$) ] $ _{{V} _{2} \rightarrow {V} _{1}}$
$W$ $=$ $nRT$ $\log _{e}{}$ $($$\dfrac { {V} _{2} - \beta n} {{V} _{1} - \beta n} $ $)$ $+$ $a{n}^{2}$$($ $\dfrac {{V} _{1} - {V} _{2}} { {V} _{1}{V} _{2}}$ $)$
Hence, the Correct Option is $'A'$.