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 modelling gases - the kinetic model ideal gases kinetic theory of gases physics

Which of the following statements is NOT a correct assumption of the model of an ideal monatomic gas?

  1. The atoms are constantly moving.

  2. The collision between atoms create pressure directly on the container of the gas.

  3. The collision between atoms are elastic.

  4. The only significant forces acting on the atoms are those that are applied as a result of collisions.

  5. The volume of each atom can be neglected.
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Pressure is created by the atom of the gas when the collide with the container. during collision the atoms apply force on the container wall. This force per unit area is the pressure of the gas.

Multiple choice modelling gases - the kinetic model ideal gases kinetic theory of gases physics

Gases exert pressure on the walls of the container because the gas molecules.

  1. Have finite volume

  2. Obey Boyle's law

  3. Possess momentum

  4. Collide with one another
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Pressure exerted by the gas molecules on the walls of container is calculated by the no. of collisions on the wall.
What a collision on the wall do is they apply a force on the wall due to the change in momentum of molecules.
$\Rightarrow$ Pressure $=\dfrac{force}{Area}=\dfrac{dP}{Adt}$
Where $dP$ is change in momentum.
Hence, the answer is Possess momentum.
Multiple choice energy efficiency energy transformations and energy transfers physics

An ideal gas is taken through a cyclic thermo dynamical process through four steps. The amounts of heat involved in these steps are ${ Q } _{ 1 }=5960 J$ ,${ Q } _{ 2 }=5585 J$, ${ Q } _{ 1 }=2980$ J,${ Q } _{ 1 }=3645 J$;espectively, The corresponding works involved are ${ W } _{ 1 }=2200 J$, ${ W } _{ 1 }= -825 J$, ${ W } _{ 2 }=-1100$ J and  ${ W } _{ 4 }$  respectively. Find The value of W and efficiency of the cycle

  1. 1315 J 10%

  2. 275 J 11%

  3. 765 J 10.82%

  4. 675 J 10.82%
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice energy efficiency energy transformations and energy transfers physics

The work done during the process when 1 mole of gas is allowed to expand freely into vacuum is:

  1. zero

  2. +ve

  3. -ve

  4. either of these
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Free expansion into a vacuum means there is no external pressure (P_ext = 0). Since work done W = integral(P_ext * dV), the work done is zero.

Multiple choice determination of atomic and isotopic mass some basic concepts of chemistry chemistry

A sample of perfect gas that initially occupies $15.0L$ at $300K$ and $1.0$ bar is compressed isothermally. To what volume must the gas be compressed to reduce its entropy by $5.0J/K$? $\left[ \ln { 0.36 } =-1.0,\ln { 2.7 } =1.0 \right] $

  1. $5.4L$
  2. $8.22L$
  3. $40.5L$
  4. $5.56L$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

For an isothermal process, delta S = nR ln(V2/V1). Since nR = PV/T = (1 bar * 15 L) / 300 K = 0.05 L*bar/K. Converting to SI units (1 bar = 10^5 Pa, 1 L = 10^-3 m^3), nR = 5 J/K. Then -5 = 5 ln(V2/15). ln(V2/15) = -1. V2/15 = e^-1 = 0.36. V2 = 15 * 0.36 = 5.4 L.

Multiple choice particles of a gas ideal gases physics

Cooking gas containers are kept in a lorry moving with uniform speed. The temperature of the gas molecules inside will :

  1. increase

  2. decrease

  3. remain same

  4. decrease for some, while increase for others
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Since the pressure and volume of gas remains same , therefore , there will not be any change in the temperature of the gas molecules.

Multiple choice laws of heat transfer heat and thermodynamics physics

Five kilomoles of oxygen is heated at constant pressure. The temperature of the oxygen gas is increased from 295 K to 305 K. If the molar heat capacity of oxygen at  constant pressure is 6.994 kcal/kmole K. The amount of heat absorbed is in kcal,

  1. 249.7

  2. 44

  3. 349.7

  4. 539.7
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Heat Q = n * Cp * deltaT. Q = 5 kmol * 6.994 kcal/kmol K * (305 - 295) K = 5 * 6.994 * 10 = 349.7 kcal.

Multiple choice chemistry changes around us can all changes be reversed? substances and objects classification of changes

Consider the expansion and contraction among solids, liquids and gases, the one which has a greater tendency to expand is :

  1. solid

  2. liquid

  3. gas

  4. none of these
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Gases have a greater tendency to expand because it has no or negligible intermolecular force of attraction between the particles. 


Hence, they can move easily, resulting in expansion whereas solid has a compact structure, they have the least tendency to expand.

Hence the correct option is C.

Multiple choice physics measurement and effects of heat thermal expansion in gases thermal expansion of fluids volume elasticity constant of gases

At constant pressure how much fraction of heat supplied to gas is converted into mechanical work ?  

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

At constant pressure, dQ = nCp dT and dW = P dV = nR dT. The fraction of heat converted to work is dW/dQ = (nR dT) / (nCp dT) = R/Cp. Since Cp = gamma*R / (gamma-1), the fraction is R / (gamma*R / (gamma-1)) = (gamma-1)/gamma.

Multiple choice physics measurement and effects of heat thermal expansion in gases thermal expansion of fluids volume elasticity constant of gases

A gas compressed to half of its volume at ${30}^{o}C$. Upto what temperature should it be heated, so that its volume increase to double of its original volume?

  1. ${60}^{o}C$
  2. $303K$
  3. $606K$
  4. $1212K$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Using P1V1/T1 = P2V2/T2. Since pressure is constant, V1/T1 = V2/T2. V1 = V, V2 = 2V. T1 = 30 + 273 = 303K. V/303 = 2V/T2. T2 = 606K.

Multiple choice physics measurement and effects of heat thermal expansion in gases thermal expansion of fluids volume elasticity constant of gases

A gas follows $VT^2 =$ const. Its volume expansion coefficient will be :-

  1. $\dfrac{2}{T}$
  2. $-\dfrac{2}{T}$
  3. $\dfrac{3}{T}$
  4. $-\dfrac{3}{T}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

$\begin{array}{l} \gamma =\dfrac { 1 }{ V } \left( { \dfrac { { dV } }{ { dT } }  } \right)  \ P{ T^{ 2 } }=cons\tan  t \ \dfrac { { nRT } }{ V } { T^{ 2 } }=cons\tan  t \ \therefore \gamma =\dfrac { 3 }{ T }  \ Hence,\, C\, the\, \, correct\, option\, .\,  \end{array}$