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 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

In a process $PT=Constant$, if molar heat capacity of a gas is $C=37.35J/mol=K$, then find the number of degrees of freedom of molecules in the gas.

  1. $n=10$
  2. $n=5$
  3. $n=6$
  4. $n=7$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a process PT = constant, the molar heat capacity is C = Cv + R / (1 - x), where PV^x = constant. Since PT = constant, P(PV/nR) = constant, so P^2V = constant, or PV^0.5 = constant. Thus x = 0.5. C = (f/2)R + R / (1 - 0.5) = (f/2)R + 2R. With C = 37.35 and R = 8.314, 37.35 = R(f/2 + 2) => 4.49 = f/2 + 2 => f/2 = 2.49 => 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

At ordinary temperatures, the molecules of a diatomic gas have only translational and rotational kinetic energies. At high temperatures, they may also have vibrational energy. As a result of this compared to lower temperatures, a diatomic gas at higher temperatures will have-

  1. lower molar heat capacity

  2. higher molar heat capacity

  3. lower isothermal compressibility

  4. higher isothermal compressibility
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Vibrational energy involves additional degrees of freedom. Thus the degrees of freedom for a diatomic gas increases at higher temperatures.

Molar heat capacity is proportional to the number of degrees of freedom of the gas.
Thus the molar heat capacity also increases for a diatomic gas at higher temperatures.

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 the temperature is increased from 0$^o$C to 273$^o$C, in what ratio the average kinetic energy of molecules changes?

  1. 1

  2. 5

  3. 4

  4. 2
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Average K.E. $= \displaystyle \frac{3}{2}$ RT
At 0$^o$C, average K.E. $= \displaystyle \frac{3}{2} \times R \times 273$
                                           $[T = (0 + 273) K]$
At 273$^o$C,    
average K.E. $= \displaystyle \frac{3}{2} \times R \times (273 + 273)$
$= \displaystyle \frac{3}{2} \times R \times 2 \times 273$
$\therefore $ Ratio = 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

When x amount of heat is given to a gas at constant pressure, it performs $\displaystyle \frac{x}{3}$ amount of work. The average number of degrees of freedom per molecule of the gas is-

  1. 3

  2. 4

  3. 5

  4. 6
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\displaystyle \frac{W}{Q}=\frac{P\Delta V}{nC _{P}\Delta T}=\frac{nR\Delta T}{nC _{P}\Delta T}=\frac{x/3}{x}$  (standard result)



$\displaystyle \Rightarrow C _{P}=3R=\left ( \frac{f}{2}+1 \right )R\Rightarrow f=4$

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 mean kinetic energy of a gas molecule is proportional to 

  1. $\displaystyle \sqrt { T } $
  2. $\displaystyle { T }^{ 3 }$
  3. $\displaystyle T$

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

The average kinetic energy of gas molecules is directly proportional to absolute temperature only; this implies that all molecular motion ceases if the temperature is reduced to absolute zero.
Hence, option C is correct.

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 diatomic gas at normal temperature is

  1. 3

  2. 4

  3. 5

  4. 6
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

In three-dimensional space, three degrees of freedom are associated with the movement of a particle. A diatomic  gas molecule thus has 6 degrees of freedom. This set may be decomposed in terms of translations, rotations, and vibrations of the molecule. The center of mass motion of the entire molecule accounts for 3 degrees of freedom. In addition, the molecule has two rotational degrees of motion and one vibrational mode The rotations occur around the two axes perpendicular to the line between the two atoms. The rotation around the atom-atom bond is not a physical rotation. At normal temp,  vibration is not possible. Hence, the total no of degrees of freedom is $f= 3+ 2=5$

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

If for a gas $\dfrac{R}{C _V}=0.67$, this gas is made up of molecules which are.

  1. Monatomic

  2. Diatomic

  3. Polyatomic

  4. Mixture of diatomic and polyatomic molecules
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For a gas, we know $\dfrac{R}{C _V}=\gamma -1$
or $0.67=\gamma -1$ or, $\gamma =1.67$
Hence the gas is monatomic.

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 average degree of freedom per molecule for a gas is 6. The gas performs 25 J of work when it expands at constant pressure. The heat absorbed by the gas is

  1. 75 J

  2. 100 J

  3. 150 J

  4. 125 J
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\Delta u=\dfrac{f}{2}RT=3RT$

$\Delta w=nR\Delta T=25.5$
$\Delta Q=\Delta V+\Delta W$
$=3nR\Delta T+nR\Delta T=4nR\Delta T$
$=4\times 25=100\ J$

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 performs Q work when it expand at constant pressure. During this process heat absorbed by the gas is 4Q. The average number of degrees of freedom for the gas is:

  1. 5

  2. 6

  3. 4

  4. 3.5
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

At constant pressure, Q = n*Cp*deltaT and W = n*R*deltaT. Given Q = 4Q (this seems to be a typo in the prompt, likely Q_heat = 4*W). If Q_heat = 4*W, then Cp*deltaT = 4*R*deltaT, so Cp = 4R. Since Cp = (f/2 + 1)R, then f/2 + 1 = 4, f/2 = 3, f = 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

N moles of an ideal diatomic gas is contained in a cylinder at temperature $T.$ On supplying some heat to cylinder, $N/3$ moles of gas disassociated into atoms while temperature remains constant. Heat supplied to the gas is

  1. $\dfrac {NRT}{3}$
  2. $\dfrac {5NRT}{2}$
  3. $\dfrac {8NRT}{3}$
  4. $\dfrac {NRT}{6}$
Reveal answer Fill a bubble to check yourself
B Correct answer
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 heat capacity at constant volume of a sample of a monoatomic gas is $35\ J/K$. Find the number of moles.

  1. $12.81 \ \ mol $
  2. $21.81 \ \ mol $
  3. $4.81 \ \ mol $
  4. $2.81 \ \ mol $
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For monoatomic gas, degrees of freedom is 3. 


Since ${ C } _{ V }=\dfrac { f }{ 2 } nR$

Hence, $35=\dfrac { 3 }{ 2 } n(8.314)$

$n=\dfrac { 70 }{ 3\times 8.314 } =2.81mol$

Answer is $2.81mol.$

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

Relation between pressure ($P$) and energy density ($E$) of an ideal gas is-

  1. $P=2/3E$
  2. $P=3/2E$
  3. $P=3/5E$
  4. $P=E$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Kinetic energy $=\dfrac{1}{2}{ MV } _{ rms }$
$\Rightarrow \dfrac{1}{2}M\left( \dfrac { 3RT }{ M }  \right) $        $[M=$ molar mass,$ { V } _{ rms }=\sqrt { \dfrac { 3KT }{ { m } }  } =\sqrt { \dfrac { 3RT }{ M }  } ]$
$=\dfrac{3}{2}RT$
$\Rightarrow K.E=\dfrac{3}{2}PV$          $[PV=RT]$
$\Rightarrow \dfrac{K.E}{V}=\dfrac{3}{2}P$
$\Rightarrow E=\dfrac{3P}{2}$        $E=$ Energy density.
Hence, the answer is $P=\dfrac{2}{3}E.$
Multiple choice physics gravitational fields representing a gravitational field gravitational field circular motion and gravitation

PRESSURE AND KINETIC INTERPRETATION OF TEMPERATURE
At what temperature the mean kinetic energy of hydrogen molecules increases to such that they will escape out of the gravitational field of earth for over?
take $({ v } _{ c }=11.2km/sec)$

  1. 12075 K

  2. 10000 K

  3. 20000 K

  4. 10075 K
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

The escape velocity is related to temperature by the formula v_rms = sqrt(3RT/M). Setting v_rms equal to the escape velocity (11.2 km/s) and solving for T gives approximately 10075 K.

Multiple choice stefan's law black body radiation heat transfer thermal properties physics

In the nuclear fusion, $ _{1}^{2}{H}+ _{1}^{3}{H}\rightarrow _{2}^{4}{He}+n$ given that the repulsive potential energy between the two nuclie is $7.7\times 10^{-14}J$, the temperature at which the gases must be heated to initiate the reaction is nearly [Boltzmann's constant $k=1.38\times 10^{-23}J/K$]-

  1. $10^{7}K$
  2. $10^{5}K$
  3. $10^{3}K$
  4. $10^{9}K$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Energy    $E \approx kT$

So,     $7.7\times 10^{-14} \approx 1.38\times 10^{-23}\times T$
$\implies \ T\approx  5.6\times 10^9 \ K$
Correct answer is option D.