Physics

Thermodynamics and Gas Laws

616 Questions

Thermodynamics and gas laws questions test the understanding of ideal gas behavior, work done during thermodynamic processes, and specific heat ratios. Key areas include isothermal, adiabatic, and isobaric expansions along with real gas deviations. These mathematical physics concepts are standard in engineering and general science competitive exams.

Ideal gas equationIsothermal and adiabatic processesThermodynamic workGas kinetic theoryReal gas behavior

Thermodynamics and Gas Laws Questions

Multiple choice chemistry basic concepts of chemistry law of multiple proportion law of multiple proportions laws of chemical combination

$CO,CO _2, C _2O _3$ follows :

  1. law of definite proportion

  2. law of multiple proportion

  3. law of conservation of mass

  4. all of the above
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The Law of Multiple Proportion: It was given by Dalton. When one element combines with the other element to form two or more different compounds, the mass of one elements, which combines with a constant mass of the other, bear a simple ratio to one another.
So, $CO,CO _2, C _2O _3$ follows law of multiple proportion.

Multiple choice chemistry basic concepts of chemistry law of multiple proportion law of multiple proportions laws of chemical combination

The molecules of nitrogen monoxide and nitrogen dioxide differ by a multiple of the mass of one oxygen. The statement can be understood by the concept of :

  1. Law of multiple proportion

  2. Nuclear fusion

  3. Van dar Waals forces

  4. Graham's Law of Diffusion(Effusion)

  5. Triple point
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The molecules of nitrogen monoxide and nitrogen dioxide differ by a multiple of the mass of one oxygen. The statement can be understood by the concept of the Law of multiple proportion.

Multiple choice physics pressure in liquids and gases common consequences of the atmospheric pressure atmospheric pressure and its consequences important points about atmospheric pressure

The pressure P of a gas and its mean translational KE per unit volume are related as:

  1. $P=\dfrac{1}{2}E$
  2. $P=E$
  3. $P=\dfrac{3}{2}E$
  4. $P=(\dfrac{2}{3})E$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\begin{array}{l} \therefore \, \, { V _{ rms } }=\sqrt { \dfrac { { 3PV } }{ M }  }  \ E=\dfrac { 1 }{ 2 } M\, \, { V _{ rms } }^{ 2 }=\dfrac { 3 }{ 2 } PV=\dfrac { 3 }{ 2 } P\, \, \left[ { \therefore \, \, V=1 } \right]  \end{array}$

Ans. (C)

Multiple choice chemistry energetics and thermochemistry bond energies and enthalpy changes bond enthalpies enthalpies for different types of reactions

Two moles of an ideal gas expended isothermally and reversibly from 1 litre to 10 litre at 300 K. The enthalpy change (in kJ) for the process is:

  1. 11.4

  2. -11.4

  3. 0

  4. 4.8
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Work done in a reversible isothermal process is:

$W = -2.303 \; nRT \; \log{\cfrac{{V} _{f}}{{V} _{i}}} ..... \left( 1 \right)$

Given:-
$n = 2 \text{ moles}$
$T = 300 \; K$
${V} _{f} = 10 \; L$
${V} _{i} = 1 \; L$
$R =$ Gas constant $= 8.314 \; {J}/{K-mol}$

Substituting these values in ${eq}^{n} \left( 1 \right)$, we have

$W = - 2.303 \times 2 \times 8.314 \times 300 \times \log{\cfrac{10}{1}}$

$\Rightarrow W = -11488.285 \; J = -11.4 \; kJ$

Now as we know that,

$\Delta{H} = \Delta{U} + W$

For an isothermal process,

$\Delta{U} = 0$

$\therefore \Delta{H} = W = -11.4 \; kJ$

Hence the enthalpy change for the given process is $-11.4 \; kJ$.

Hence, the correct option is $\text{B}$
Multiple choice maths the trapezium rule approximation errors and approximations the need for approximation

The pressure P and volume V of a gas are connected by the relation $PV^{1/4}=constant$. The percentage increase in the pressure corresponding to a deminition of $\dfrac12 \%$ in the volume is

  1. $\dfrac {1}{2}$ %
  2. $\dfrac {1}{4}$ %
  3. $\dfrac {1}{8}$ %

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
$PV^{1/4}=constant$
$\displaystyle \Rightarrow P=\dfrac{k}{V^{{1}/{4}}}$
$\displaystyle \Rightarrow \dfrac{dP}{dV}=-\dfrac{k}{4}V^{-{5}/{4}}$
Percentage error in V $\displaystyle= -\dfrac{1}{2}\%$
$\Rightarrow\displaystyle \dfrac{\Delta V}{V} =-\dfrac{1}{200}$
$\Rightarrow\displaystyle {\Delta V}=-\dfrac{V}{200}$
Approximate change in $P\displaystyle=dP=(\dfrac{dP}{dV}){\Delta V}$
                                           $\displaystyle  =\dfrac{1}{800} {kV^{-1/4}}=\dfrac{1}{8}\%$ of P
Percentage increase in  $V \ \displaystyle =\dfrac{1}{8}\%$
Multiple choice separation of components of mixtures methods of separation elements, compounds and mixtures chemistry

Gaseous diffusion works on the principle that

  1. Molecules of a lighter isotope would pass through a porous barrier more readily than those of a heavier isotope

  2. Molecules of a heavier isotope would pass through a porous barrier more readily than those of a lighter isotope

  3. Heavier and lighter particles will pass readily through the porous barrier

  4. None of above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Gaseous diffusion works on the principle that molecules of a lighter isotope would pass through a porous barrier more readily than those of a heavier isotope.

Multiple choice evs - i substances, objects and energy renewable resources alternative fuels and energy sources alternative sources of energy

A sealed container with negligible coefficient of volumetric expansion contains helium (a mono-atomic gas). When it is heated from 300 K to 600 K, the average KE of helium atoms is:

  1. halved

  2. unchanged

  3. doubled
  4. increased by factor $\sqrt2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$the\quad avarage\quad KE\quad of\quad n\quad moles\quad of\quad gas\quad at\quad temp\quad T\quad is\quad \dfrac { 3 }{ 2 } nRT\ when\quad temp\quad is\quad doubled\quad \dfrac { 3 }{ 2 } nR(2T)=2\times \dfrac { 3 }{ 2 } nRT\ so\quad KE\quad doubles$

Multiple choice various types of barometer fluids physics

If the compressibility of water is $\sigma$ per unit atmospheric pressure, then the decrease in volume $V$ due to atmospheric pressure $p$ will be

  1. $\sigma p/V$
  2. $\sigma pV$
  3. $\sigma /pV$
  4. $\sigma V/p$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Compressibility sigma = -(1/V) * (dV/dp). Therefore, dV = -sigma * V * dp. For a pressure change p, the magnitude of volume decrease is sigma * V * p.

Multiple choice various types of barometer fluids physics

In a certain region of space, there are n molecules per unit volume. The temperature of the gas T. The pressure of the gas will be: 

  1. $nRT$
  2. $nkT$
  3. $\dfrac{{nT}}{k}$
  4. $\dfrac{{nT}}{R}$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The ideal gas law is PV = nRT, where n is the number of moles. In terms of number of molecules N, it is PV = NkT, where k is the Boltzmann constant. Dividing by volume V gives P = (N/V)kT, where N/V is the number density n, leading to P = nkT.

Multiple choice physics propagation of sound waves sound as a wave of disturbance vibrations in a tuning fork vibrations in tuning fork

What is the ratio of the speed of sound in neon and water vapor at the same temperature. It is nearest to :

  1. 2.5

  2. 2

  3. 1.5

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

According to the Avogadro's law , at same temperature and pressure the number of molecules of different gasses are equal. Therefore density will not change in either case of neon or water vapor. And the speed of sound waves is a function of density and temperature. So in this case doesn't matter.
Option "D" is correct. 

Multiple choice evs let's play with water law of constant proportion - i law of constant proportion law of definite proportions

Law of definite proportion does not apply to nitrogen oxide because:

  1. Atomic weight of nitrogen is not conserved

  2. Molecular weight of nitrogen is variable

  3. Equivalent weight of nitrogen is variable

  4. Atomic weight of oxygen is variable
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Nitrogen forms multiple compounds with oxygen. Hence, we cannot apply the law of definite proportion to nitrogen oxide.

Multiple choice physics solar equipment solar power plant production of electricity from solar energy solar power solar energy and its applications generation of electricity

Relation between constant volume specific heat $(C _v)$ and degree of freedom $(f)$ for a gas is ($R$ is gas constant) 

  1. $C _v=\dfrac{f}{2} R$
  2. $C _v=fR$
  3. $C _v=\dfrac{f}{R}$
  4. $C _v=\dfrac{R}{f}$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The internal energy of 1 mole of an ideal gas at temperature $T$ , having $f$ degrees of freedom is given by ,

          $U=N _{A}\times(1/2)fk _{B}T=(1/2)fRT$  ......................eq1
where $N _{A}=$Avogadro's number
           $k _{B}=$ Boltzmann's constant
from first law of thermodynamics ,
           $dU=dQ-dW$  ............eq2
At constant volume , $dV=0$ 
hence  $dW=PdV=0$
eq2 becomes , $dU=dQ$
but       $dQ=C _{V}dT$  (for 1 mole of gas)
therefore $dU=C _{V}dT$
or            $C _{V}=dU/dT$
putting the value of $U$ in this equation, we get
                $C _{V}=\frac{d(1/2)fRT}{dT}=\frac{f}{2}R$

Multiple choice physics solar equipment solar power plant production of electricity from solar energy solar power solar energy and its applications generation of electricity

Temperature of a gas is $20^o$C and pressure is changed from $1.01\times 10^5$ Pa to $1.165\times 10^5$ Pa. If volume is decreased isothermally by $10\%$. Bulk modulus of gas is?

  1. $1.55\times 10^5$
  2. $0.155\times 10^5$
  3. $1.4\times 10^5$
  4. $1.01\times 10^5$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
$B=-V\dfrac{\Delta P}{\Delta V}$
$=V _0\dfrac{(1.165-1.01)\times 10^5}{0.1\ V}$
$=1.55\times 10^5$
Multiple choice physics nuclei beta decay change in nucleus due to radioactive decay alpha, beta and gamma particles (rays) and their properties

A certain mass of an ideal diatomic gas contained in a closed vessel is heated. It is observed that half the amount of gets dissociated, but the temperature remains constant. The ratio of the heat supplied to the gas to the initial internal energy of the gas will be

  1. $1:2$
  2. $1:4$
  3. $1:5$
  4. $1:10$
Reveal answer Fill a bubble to check yourself
A Correct answer