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 life cycle of stars evolution and end stages of stars the stars stars

Which of the following statements about degeneracy pressure is not true?

  1. Degeneracy pressure varies with the temperature of the star.

  2. Degeneracy pressure can halt gravitational contraction of a star even when no fusion is occurring in the core.

  3. Degeneracy pressure keeps any protostar less than 0.08 solar mass from becoming a true, hydrogen-fusing star.

  4. Degeneracy pressure supports white dwarfs against gravity.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Degeneracy pressure is the pressure which prevent the neutron star from becoming the Black hole.

Degeneracy pressure does not varies with the temperature of the star.

Multiple choice viscosity option b: engineering physics properties of matter physics

Viscosity of the fluids is analogous to:

  1. Random motion of the gas molecules

  2. Friction between the solid surfaces

  3. integral motion

  4. Nonuniform motion of solids
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Viscosity is the property by virtue of which a liquid offers resistance for the relative motion between its layer.
Friction is the force between two solid surfaces which offers resistance for the relative motion between them. 
So, viscosity of the fluids is analogous to friction between the solid surfaces.

Multiple choice physics oscillations introduction to sound free, forced and damped oscillations resonance

An ideal gas enclosed in a vertical cylindrical container supports a freely moving piston of mass M. The piston and the cylinder have equal cross sectional area A. When the piston is in equilibrium, the volume of the gas $ \mathrm{V} _{0}  $ and its pressure is $  \mathrm{P} _{0} $ The piston is slightly displaced from the equilibrium position and released. Assuming that the system is completely isolated from its surrounding, the piston executes a simple harmonic motion with frequency.

  1. $ \dfrac{1}{2 \pi} \dfrac{\mathrm{A} \gamma P _{0}}{V _{0} M} $
  2. $ \dfrac{1}{2 \pi} \dfrac{V _{0} M P _{0}}{A^{2} \gamma} $
  3. $ \dfrac{1}{2 \pi} \sqrt{\dfrac{A^{2} \gamma P _{0}}{M V _{0}}} $
  4. $ \dfrac{1}{2 \pi} \sqrt{\dfrac{M V _{0}}{A \gamma P _{0}}} $
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For an adiabatic process, PV^gamma = constant. The restoring force for a small displacement x is F = -A * dP = -A * (gamma * P0 / V0) * (A * x). This leads to the SHM equation with omega^2 = (A^2 * gamma * P0) / (M * V0). Frequency f = omega / (2 * pi).

Multiple choice chemistry chemical equilibrium equilibrium in chemical processes introduction to equilibrium chemical equilibrium and acids-bases

Attainment of equilibrium in a coloured gaseous reversible reaction is detected by the constancy of:

  1. colour

  2. density

  3. pressure

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

The intensity of color represents the concentration of either reactant or product. Thus when in a colored gaseous reversible reaction, the color has attained constant intensity, the concentrations of reactants and products have reached equilibrium values. In other words, an equilibrium is attained.


Hence, the correct option is A.

Multiple choice chemistry chemical equilibrium equilibrium in chemical processes introduction to equilibrium chemical equilibrium and acids-bases

A mixture of three gases P (density 0.90), Q (density 0.178) and R (density 0.42) is enclosed in a vessel at the constant temperature. When the equilibrium is established:

  1. the gas P will be at the top of the vessel

  2. the gas Q will be at the top of the vessel

  3. the gas R will be at the top of the vessel

  4. the gases will mix homogeneously throughout the vessel.
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Independent of density of gases, in equilibrium gases will mix homogenously, this comes from the fact that gases occupies entire volume of container. This also can be thought as gases tries to reduce energy. Hence they separate as far as possible.

Multiple choice chemistry enthalpy changes enthalpy changes and enthalpy profile diagrams enthalpy study of enthalpy

Heat of reaction at constant pressure and heat of reaction at constant volume for the gaseous reaction $N _2 + 3H _2 \longrightarrow  2NH _3$ differ $(\Delta H- \Delta U)$ by the amount:

  1. $2RT$
  2. $-2RT$
  3. $3RT$
  4. $RT$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The heat of reaction at constant pressure and heat of reaction at constant volume for the gaseous reaction 


$N _2 + 3H _2 \longrightarrow  2NH _3$

$\Delta H = \Delta U + (\Delta n _g)\times RT$

$\Delta H = \Delta U + (-2)\times RT$

This is because the change in the number of moles of gaseous products and the gaseous reactants in the above reaction is -2.

$\Delta H - \Delta U = (-2)\times RT$

Hence, option B is correct.

Multiple choice chemistry enthalpy changes enthalpy changes and enthalpy profile diagrams enthalpy study of enthalpy

Select the correct option(s):

  1. $q=nC _{v}\mathrm{d} T$ is applicable to all substances during heating/cooling at constant 'v'.
  2. $q=nC _{v}\mathrm{d} T$ is applicable to ideal gas during heating/cooling at constant 'v'.
  3. $\mathrm{d} U=nC _{v}\mathrm{d} T$ is applicable for real gas at constant 'v'
  4. $\mathrm{d} U=nC _{v}\mathrm{d} T$ is applicable for ideal gas at constant 'v' only
Reveal answer Fill a bubble to check yourself
A,B Correct answer
Explanation

An isochoric process is a thermodynamic process during which the volume of the closed system undergoing such a process remains constant. 


Work done by system is given as $W=P\Delta V$

So workdone by the ideal and real gases will be zero in all cases.

From first law of thermodynamics, $Q=W+\Delta U$
$\Longrightarrow Q=\Delta U$

$\Delta U=n{ C } _{ v }\Delta T$ is applicable for all conditions whether the volume is constant or not.

Multiple choice chemistry enthalpy changes enthalpy changes and enthalpy profile diagrams enthalpy study of enthalpy

A mixture of 2 mole of carbon monoxide and one mole of oxygen in a closed vessel is ignited to get carbon dioxide. If $\Delta H$ is the enthalpy change and $\Delta U$ is the change in internal energy, then:

  1. $\Delta H >\Delta U$
  2. $\Delta H <\Delta U$
  3. $\Delta H =\Delta U$

  4. can't be predicted
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

A mixture of 2 moles of carbon monoxide and one mole of oxygen in a closed vessel is ignited to get carbon dioxide. 


 $2CO(g)+ O _2(g) \longrightarrow 2CO _2(g)$

 $\Delta n _g =2-3=-1$

If $\Delta H$ is the enthalpy change and $\Delta U$ is the change in internal energy, then:

 $\Delta H=\Delta U+\Delta n _gRT=\Delta U-RT$

Hence,  $\Delta H <\Delta U$

Multiple choice physics heat engine: second law of thermodynamics second law of thermodynamics the second law of thermodynamics second law of thermodynamic

For a gas $\cfrac{R}{C _{p}}=0.4$. For this gas calculate the following-

  1. atomicity and degree of fredom
  2. value of $C _{v}$ and $\gamma$
  3. mean gram-molecular kinetic energy at $300\ k$ temperature

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

Given R/Cp = 0.4, we find Cp = R/0.4 = 2.5R. Since Cp - Cv = R, we get Cv = 1.5R. The ratio of specific heats γ = Cp/Cv = 2.5R/1.5R = 5/3 = 1.67. This corresponds to a monatomic gas with f = 3 degrees of freedom. Option B correctly asks for Cv and γ values, which are the natural quantities to calculate from the given.

Multiple choice physics heat engine: second law of thermodynamics second law of thermodynamics the second law of thermodynamics second law of thermodynamic

A new soft drink bottle is opened, allowing gas to escape into the atmosphere. As the gas escapes, its degree of disorder increases. Identify by which of the following law this can be explained ?

  1. First law of thermodynamics (conservation of energy)

  2. Second law of thermodynamics (law of entropy)

  3. Ideal gas law

  4. Heat of fusion and heat of vaporization equation

  5. Heat engine efficiency
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

The second law of thermodynamics states that the total entropy of an isolated system always increases over time, or remains constant in ideal cases where the system is in a steady state or undergoing a reversible process. Thus an opened bottle allows gas to escape since the entropy increases(disorderliness increases) over time.

Multiple choice physics heat engine: second law of thermodynamics second law of thermodynamics the second law of thermodynamics second law of thermodynamic

The work done in heating on emole of an ideal gas at constant pressure from ${ 15 }^{ 0 }C\quad to\quad { 25 }^{ 0 }C$ is

  1. 1.987 cal

  2. 198.7 cal

  3. 9.935 cal

  4. 19.87 cal
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

Work done at constant pressure W = n * R * deltaT. n = 1 mole, R = 1.987 cal/mol*K, deltaT = 25 - 15 = 10 K. W = 1 * 1.987 * 10 = 19.87 cal.

Multiple choice physics heat engine: second law of thermodynamics second law of thermodynamics the second law of thermodynamics second law of thermodynamic

Heat is supplied to a diatomic gas at constant pressure. The ratio of $\Delta Q:$$\Delta$U:$\Delta$W is:

  1. 5:3:2

  2. 7:5:2

  3. 2:3:5

  4. 2:5:7
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
For a diatomic gas,

$C _V=\dfrac 52R$

$C _P=\dfrac 72 R$

We know that
$\therefore \Delta Q=nC _P \Delta T=n (\dfrac 72 R)\Delta T$

$\Delta U=nC _V \Delta T=n(\dfrac 52 R)\Delta T$

According to first law of thermodynamics,

$\Delta W=\Delta Q-\Delta U=nR \Delta T$

$\therefore \Delta Q: \Delta U: \Delta W=7:5:2$