Physics

Thermodynamics and Gas Laws

626 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 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

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 $

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

An ideal gas having initial pressure P, volume V and temperature T is allowed to expand adiabatically until its volume becomes $5.66$V while its temperature falls to $T/2$. How many degrees of freedom do the gas molecules have?

  1. 7
  2. $5$.

  3. 6

  4. 8
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Adiabatic equation of perfect gas is given as $TV^{r-1}=$ constant
$m=T _{1}V _{1}^{(r-1)}=T _{2}V _{2}^{(r-1)}$
$T _{1}=T _{1}V _{1}=V _{1}V _{2}=5.66\ V$
and $T _{2}=\dfrac{T}{2}$
$TV^{r-1}=\dfrac{T}{2}(5.66\ V)^{r-1}$
$2=5.66^{r-1}$
Taking $\log$ on both sides
$(r-1)\log 5.66=\log 2(r-1)$
$r=\dfrac{\log 2}{\log 5.66}=1+0.3010/0.75$
$r=1+0.4$
$r=1.4$ for $r=1.4$ Agree of freedom $=5$
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

A gas has molar heat capacity $C = 4.5\ R$ in the process $PT = constant$. Find the number of degrees of freedom (n) of molecules in the gas.

  1. $n = 7$
  2. $n = 3$
  3. $n = 5$
  4. $n = 2$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For a process PT = constant, P * (PV/nR) = constant, so P^2 * V = constant. This is a polytropic process PV^x = constant with x = 1/2. Molar heat capacity C = R/(gamma-1) + R/(1-x). 4.5R = R/(2/f) + R/(1-0.5) = fR/2 + 2R. 4.5 = f/2 + 2, so f/2 = 2.5, f = 5.

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

A gas undergoes a process such that $P \alpha \dfrac{1}{T}$. If the molar heat capacity for this process is $24.93 \,J/mol \,K$, then what is the degree of freedom of the molecules of the gas?

  1. $8$
  2. $4$
  3. $2$
  4. $6$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Given P proportional to 1/T, using the ideal gas law PV = nRT, this implies V is proportional to 1/T^2 or VT^2 = constant. The molar heat capacity C = Cv + R/(1 - x) where V proportional to T^-x, giving x = -2. Solving this yields Cv and subsequently the degrees of freedom f = 2.

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 degrees of freedom of a triatomic gas is? (consider moderate temperature)

  1. $6$
  2. $4$
  3. $2$
  4. $8$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The general epression for degree of freedom is $DOF=3N-n$

here, DOF means degree of freedom, N is number of particle, and n is the number of holonomic constraints.
for a triatomic molecule, the number of particle is 3 and since the separation between three atoms are fixed so, the number of constraints is 3.
hence, $DOF=(3\times 3)-3$
$DOF=9-3$
$DOF=6$

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

For gas, if the ratio of specific heats at constants pressure $P$ and constant volume $V$ is $\gamma $, then the value of degree of freedom is:

  1. $\dfrac{\gamma +1}{\gamma -1}$
  2. $\dfrac{\gamma -1}{\gamma +1}$
  3. $\dfrac{1}{2}(\gamma-1)$
  4. $\dfrac{2}{\gamma-1}$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The ratio of specific heats is given by gamma = Cp / Cv. Using Mayer's relation Cp - Cv = R and Cv = fR / 2, we can express gamma as 1 + (2 / f). Rearranging this formula yields the degree of freedom f equal to 2 / (gamma - 1).

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

When an ideal monoatomic  gas is heated zt constant pressure , which of the following may be true

  1. $\dfrac {dU}{dQ} = \frac {3}{5}$
  2. $\dfrac {dW}{dQ} = \frac {2]}{5}$
  3. $\dfrac {dU}{dQ} = \frac {4}{5}$
  4. $dW + dU = dQ $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

From First Law: dQ = dU + dW. This is always true for any process. For monatomic gas at constant pressure, dU = (3/2)nRdT and dW = PdV = nRdT, so dQ = (5/2)nRdT. Then dU/dQ = 3/5 and dW/dQ = 2/5, not 4/5. Option D is the fundamental First Law statement.

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

If $\gamma $ be the ration of specific heats of a perfect gas, the number of degree of freedom of a molecule of the gas is:

  1. $\dfrac{{25}}{2}\left( {\gamma - 1} \right)$
  2. $\dfrac{{3\gamma - 1}}{{2\gamma - 1}}$
  3. $\dfrac{2}{{\gamma - 1}}$
  4. $\dfrac{9}{2}(\gamma - 1)$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The relationship between the adiabatic index gamma and degrees of freedom f is gamma = 1 + 2/f. Solving for f gives f = 2 / (gamma - 1).