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 speed of sound in gas speed of a travelling wave oscillation and waves waves physics

With increase in temperature, the rms speed and wave speed in a gas

  1. increases with temperature

  2. decreases with temperature

  3. are independent of temperature

  4. none of the above
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Both RMS speed and speed of sound in gas are directly proportional to temperature. Thus, both the speeds increases with temperature

The correct option is (a)

Multiple choice speed of sound in gas speed of a travelling wave oscillation and waves waves physics

The velocity of sound at the same pressure in two monoatomic gases of densities $ \rho _1$  and $\rho _2$ are $v _1$ and $v _2 $ respectively. If $ \dfrac {\rho _1}{\rho _2} = 4 $ then the value of $ \dfrac {v _1}{v _2} $ is:-

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

Velocity v = sqrt(gamma * P / rho). Since pressure P and gamma are constant, v is inversely proportional to sqrt(rho). Thus, v1 / v2 = sqrt(rho2 / rho1). Given rho1 / rho2 = 4, then rho2 / rho1 = 1/4. So v1 / v2 = sqrt(1/4) = 1/2.

Multiple choice physics pressure in liquids and gases pressure dependence on force and area concept of pressure pressure on surface

A same amount of same gas of temperature T are enclosed in a three identical vessel A,B, & C. The temperature of wall of three container is $T _A , T _B$ & $T _C (T _A > T _B > T _C)$ respectively. The pressure on wall of vessel. 

  1. $P _A > P _B > P _C$
  2. $P _A < P _B < P _C$
  3. $P _A = P _B = P _C$

  4. Deta's are insufficient to decide.
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$\begin{array}{l} PV=nRT \ P\propto T \ \therefore \, { T _{ A } }>{ T _{ B } }>{ T _{ C } } \ So, \ { P _{ A } }>{ P _{ B } }>{ P _{ C } } \ Hence, \ option\, \, A\, \, iscorrect\, answer. \end{array}$

Multiple choice physics pressure in liquids and gases pressure dependence on force and area concept of pressure pressure on surface

Air is pumped into an automobile tube upto a pressure of $200 \mathrm{kP}  $ in the morning when the air temperature is $  22^{\circ} \mathrm{C}  $ . During the day, temperature rises to $ 42^{\circ} \mathrm{C}  $ and the tube expands by $2 \%  $ . The pressure of the air in the tube at this temperature, will be approximately

  1. $212 \mathrm{kPa} $
  2. $209 k P a $
  3. $206 \mathrm{kPa} $
  4. $200 \mathrm{kPa} $
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$\begin{array}{l} Given\, \, { P _{ 1 } }=200\, \, k\, pa \ { T _{ 1 } }=22^{ \circ  }C=273+22-295K \ { T _{ 2 } }=42^{ \circ  }C=273+42=315K \ PV=nRT \ \left( { \dfrac { { PV } }{ T }  } \right) ={ { constant } }\left[ \begin{array}{l} { { Since } }\, \, number\, \, of\, \, moles \ and\, \, R\, \, are\, \, { { constant } } \end{array} \right]  \ \dfrac { { { P _{ 1 } }{ V _{ 1 } } } }{ { { T _{ 1 } } } } =\dfrac { { { P _{ 2 } }{ V _{ 2 } } } }{ { { T _{ 2 } } } }  \ \dfrac { { 200\times 1 } }{ { 295 } } =\dfrac { { { P _{ 2 } }\left( { 1.02 } \right)  } }{ { 315 } }  \ { P _{ 2 } }=\dfrac { { 200\times 315 } }{ { 295\times 1.02 } } \, \, kPa \ = 209\, \, kPa \end{array}$

Option B.

Multiple choice chemistry study of the first element - hydrogen position of hydrogen in the periodic table hydrogen in periodic table trends in groups and periods

At the temperature of liquefaction of air, the ratio of ortho and para hydrogen is:

  1. 1:1

  2. 1:3

  3. 3:1

  4. 3:2
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

At room temperature, the ratio of ortho to para hydrogen is 3:1. As the temperature decreases toward the liquefaction point, the equilibrium shifts toward para hydrogen, reaching 1:1 at the liquefaction temperature.

Multiple choice physics heat - measurement introduction to temperature application of various thermometric scales different types of thermometers

A container having some gas was kept in a moving train. The temperature of the gas in the container will be

  1. Increases slightly

  2. Remain the same

  3. Decrease

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

The correct answer is option(B).

When we say that the temperature of a gas is the measure of the kinetic energy of the gas molecules, we have to find the velocities of the gas molecules in the centre of mass frame of the gas.
If we chose a different inertial frame then the temperature doesn't increase or decrease. 

Multiple choice physics the kinetic model of matter gases and the kinetic theory concept of ideal gas and state equation of ideal gas behaviour of perfect gas and kinetic theory of gases

Two identical containers A and B with frictionless pistons contain the same ideal gas at the same temperature and same volume 'V'. The mass of the gas in A is $m _{A}$ and that in B is '$m _{B}$' . The gas in each cylinder is now allowed to expand isothermally to the final volume 2V. The changes in the pressure in A and B are found to be $\Delta P$ and 1.5$\Delta P$ respectively. Then

  1. $4m _{A} = 9m _{B}$
  2. $2m _{A} = 3m _{B}$
  3. $3m _{A} = 2m _{B}$
  4. $9m _{A} = 4m _{B}$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation
Since it is isothermal, the Temperature T is constant and as both the gases are expanded to the same volume 2V, in the gas equation PV=nRT, V cancels out.
$\dfrac { P }{ m } =Constant\\ \dfrac { \Delta p }{ { m } _{ A } } =\dfrac { 1.5\Delta p }{ { m } _{ B } } \\ \therefore { m } _{ B }=1.5{ m } _{ A }\\ \therefore 2{ m } _{ B }=3{ m } _{ A }$
Multiple choice physics the kinetic model of matter gases and the kinetic theory concept of ideal gas and state equation of ideal gas behaviour of perfect gas and kinetic theory of gases

What should be the percentage increase in the pressure so that the volume of a gas may decrease by 5% at constant teperature.

  1. 5%

  2. 10%

  3. 5.26%

  4. 4.26%
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

If initial volume = v, final volume = $\dfrac{95v}{100}$
If P1 = initial pressure, P2 (final pressure) =$\dfrac{100P _1}{95}$
So increase in pressure = $\dfrac{100P _1}{95 - P _1} = \dfrac{5P _1}{95}$
So, percentage increase in pressure = $5\times \dfrac{100}{95} = 5.26%$

Multiple choice physics the kinetic model of matter gases and the kinetic theory concept of ideal gas and state equation of ideal gas behaviour of perfect gas and kinetic theory of gases

A given mass of ideal gas has volume (V) at pressure (P) and the room temperature. If its pressure is first increased by 50% and then decreased by 50% (both at constant temperature only), the volume becomes.

  1. 4V/3

  2. 3V/4

  3. V

  4. 4V/5
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

${ P } _{ 1 }{ V } _{ 1 }={ P } _{ 2 }{ V } _{ 2 }\ First\quad increased\quad by\quad 50\quad i.e\quad { P } _{ 2 }=1.5{ P } _{ 1 }\ { V } _{ 2 }=\dfrac { 1 }{ 1.5 } { V } _{ 1 }\ Now\quad decreased\quad by\quad 50\quad i.e\quad { P } _{ 3 }=\dfrac { { P } _{ 2 } }{ 2 } =\dfrac { 1.5 }{ 2 } { P } _{ 1 }\ { P } _{ 2 }{ V } _{ 2 }={ P } _{ 3 }V _{ 3 }\ V _{ 3 }=2{ V } _{ 2 }\ V _{ 3 }=\dfrac { 2 }{ 1.5 } { V } _{ 1 }\ V _{ 3 }=\dfrac { 20 }{ 15 } { V } _{ 1 }=\dfrac { 4 }{ 3 } { V } _{ 1 }$

Multiple choice physics the kinetic model of matter gases and the kinetic theory concept of ideal gas and state equation of ideal gas behaviour of perfect gas and kinetic theory of gases

Boyle's law is applicable when

a) temperature is constant

b) gas is at high temperature and low pressure

c) the vessel enclosing the gas is good conductor

d) the process is isothermal

  1. a & b

  2. b,c & d

  3. a,b & c

  4. a,b,c & d
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

(a) Temperature must be constant to apply boyle's law
(b)From vanderwaal's equation when non idealities of a gas is undertaken then we can apply boyle's only when temperture is high and pressure is low
(c)Vessel enclosing must be a good conductor so there is no any possibilities of adiabatic process
(d) Temperature must be constant therefore, process must be isothermal
Hence all are correct
Hence option(D)

Multiple choice physics save energy types of sources of energy renewable and non-renewable resources renewable and non-renewable sources of energy

As steam expands in turbine-

  1. its pressure increases

  2. its specific volume increases

  3. its boiling point increases

  4. its temperature increases
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

 Specific volume is defined as the number of cubic meters occupied by one kilogram of a particular substance. When steam expands its volume increases hence its specific volume increases.