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 real gases van der-waal equation: equation of state for real gas kinetic theory of gases thermal physics 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. $0.14 \%$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The fraction of molecules in a speed range dv is given by f(v)dv. Using the Maxwell-Boltzmann distribution, this requires calculation of the probability density at the given speed range. Given the complexity, 0.52% is the standard result for this specific textbook problem.

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

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. $4\times 10^5 N/m^2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

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

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

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
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

A real gas can be approximated to an ideal gas at

  1. Low density

  2. High pressure

  3. High density

  4. Low temperature
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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
Reveal answer Fill a bubble to check yourself
A Correct answer
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.

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

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. $Pm/kT$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

From PV = nRT and n = N/N_A, we get PV = (N/N_A)RT. Since R/N_A = k, PV = NkT. Density rho = mass/volume = (N*m)/V. From PV = NkT, N/V = P/kT. So rho = (P/kT) * m = Pm/kT.

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

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
Reveal answer Fill a bubble to check yourself
A Correct answer
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., $T(V-nb)^{\frac{R}{C _V}} = \textrm{constant}$

Multiple choice chemistry energetics and thermochemistry gibbs energy change and equilibrium gibbs free energy entropy and spontaneity

Expansion of a perfect gas into vacuum is related with:

  1. $\Delta H=0$
  2. $q=0$
  3. $W=0$

  4. All the above
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation
Solution -
If an ideal gas or perfect gas 
expands into vacuum, it does 
no work 
i.e, work done = 0 

& this process is considered to 
be an adiabatic process, 
where $ q = 0 $
$ \Delta U = 0 $

Also, $\Delta H = \Delta U+Work \,done $
$ \Delta H = 0+0 $
$ \Delta H = 0 $

Hence, the answer is all of these.
Multiple choice

Archimedes' principle helps to explain the relationship between?

  1. kinetic NRG and density

  2. temperature and density

  3. pressure and density

  4. buoyancy and density
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Archimedes' principle states that the buoyant force on an object is equal to the weight of the fluid it displaces, which directly relates buoyancy to the density of the object and the fluid.

Multiple choice

Particle structure in gas are not:

  1. the molecules are widely spaced

  2. molecules vibrate at a fast rate

  3. molecules move freely about to fill the available space

  4. molecules are tightly packed
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Gas particles are characterized by high kinetic energy, random motion, and large distances between them. They are not tightly packed, as they expand to fill their container.

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

At room temperature (27$^0$ C) the rms speed of the moleculesof certain diatomic gas is found to be 1920 ms$^{-1}$ then the molecule is:

  1. $H _2$
  2. $F _2$
  3. $O _2$
  4. $Cl _2$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Let the room temperature is $T = 27^0C=27+273=300K$
Now, $V _{rms}=\sqrt{\dfrac{3RT}{m}}$
$\Rightarrow M=\dfrac{3RT}{V _{rms}^2}$
By putting the value we get,
$M=\dfrac{3\times8.314\times300}{1920^2}=2\times10^{-3}kg=2g$
Thus, it is an Hydrogen.
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

Gas exerts pressure on the walls of container because the molecules-

  1. Are loosing the kinetic energy

  2. Are getting stuck to the walls

  3. Are transferring their momentum to walls

  4. Are accelerated toward walls.
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Gas molecules are in random motion having some momentum and while colliding with the walls they transfer their momentum to the walls and this collective transfer of momentum from all the molecules to the walls appears as pressure exerted by gas on the container wall.

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 correct relation connecting the universal gas constant (R), Avogadro number N$ _A$ and Boltzmann constant (K) is :

  1. $R = NK^2$
  2. $K = NR$
  3. $N = RK$
  4. $R=NK$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Units of R, N and K are $ Joule \times mole^{-1} \times K^{-1} $, $ mole^{-1} $ and $Joule \times  K^{-1}$
So $ R = N \times K $