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 principal and molar specific heats of gases isothermal and adiabatic processes specific heat capacity heat and thermodynamics physics

${C} _{P}$ and ${C} _{V}$ are specific heats at constant pressure and constant volume respectively. It is observed that
${C} _{P}-{C} _{V}=a$ for hydrogen gas
${C} _{P}-{C} _{V}=b$ for nitrogen gas
The correct relation between $a$ and $b$ is then

  1. $a=28b$
  2. $a=\cfrac{1}{14}b$
  3. $a=b$
  4. $a=14b$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For ideal gas
${C} _{P}-{C} _{V}=R/M$
If ${C} _{P}$ and ${C} _{V}$ are specific heats $\left( J/kg- _{  }^{ o }{ C } \right) $
$M=$ molar mass of gas
$\Rightarrow a=R/2$ and $b=R/28$
$\Rightarrow$ $a=14b$

Multiple choice principal and molar specific heats of gases isothermal and adiabatic processes specific heat capacity heat and thermodynamics physics

For hydrogen gas $C _{p}-C _{v}=a$ and for Oxygen gas $C _{p}-C _{v}=b $, where $C _{p}$ and $C _{v}$ are molar specific heats. Then the relation between a and b. is

  1. a $=$ 16b
  2. b $=$ 16a
  3. a $=$ 14b
  4. a $=$ b
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

For any ideal gas,$C _p-C _v=nR$, where $R$ is the gas constant.
That is $C _p-C _v$ per mole for any gas is a constant value.
So, $a=b$
Option D.

Multiple choice principal and molar specific heats of gases isothermal and adiabatic processes specific heat capacity heat and thermodynamics physics

Three perfect gases at absolute temperatures ${T} _{1},{T} _{2}$ and ${T} _{3}$ are mixed. The masses of molecules are ${m} _{1},{m} _{2}$ and ${m} _{3}$ and the number of molecules are ${n} _{1},{n} _{2}$ and ${n} _{3}$ respectively. Assuming no loss of energy, the final temperature of the mixture is:

  1. $\cfrac { { n } _{ 1 }{ T } _{ 1 }+{ n } _{ 2 }{ T } _{ 2 }+{ n } _{ 3 }{ T } _{ 3 } }{ { n } _{ 1 }+{ n } _{ 2 }+{ n } _{ 3 } } $
  2. $\cfrac { { n } _{ 1 }{ T } _{ 1 }+{ n } _{ 2 }{ { T } _{ 2 } }^{ 2 }+{ n } _{ 3 }{ { T } _{ 3 } }^{ 2 } }{ { n } _{ 1 }{ T } _{ 1 }+{ n } _{ 2 }{ T } _{ 2 }+{ n } _{ 3 }{ T } _{ 3 } } $
  3. $\cfrac { { n } _{ 1 }{ { T } _{ 1 } }^{ 2 }+{ n } _{ 2 }{ { T } _{ 2 } }^{ 2 }+{ n } _{ 3 }{ { T } _{ 3 } }^{ 2 } }{ { n } _{ 1 }{ T } _{ 1 }+{ n } _{ 2 }{ T } _{ 2 }+{ n } _{ 3 }{ T } _{ 3 } } $
  4. $\cfrac { \left( { T } _{ 1 }+{ T } _{ 2 }+{ T } _{ 3 } \right) }{ 3 } $
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

The final temperature of a mixture of gases is the weighted average of the temperatures, where the weights are the number of molecules (or moles) of each gas.

Multiple choice principal and molar specific heats of gases isothermal and adiabatic processes specific heat capacity heat and thermodynamics physics

For hydrogen gas $C _{p} -C _{v} = a$ and for oxygen gas $C _{p} - C _{v}=b$, where $C _{p}$ and $C _{v}$ are molar specific heats. Then the relation between 'a' and 'b' is

  1. $a=16b$
  2. $b=16a$
  3. $a=4b$
  4. $a =b$
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

According to Mayer's relation, the difference between molar specific heats at constant pressure and constant volume is equal to the universal gas constant R for any ideal gas, meaning Cp - Cv = R for both hydrogen and oxygen. Thus, a equals b.

Multiple choice principal and molar specific heats of gases isothermal and adiabatic processes specific heat capacity heat and thermodynamics physics

The specific heat of air at constant pressure is $1.005\ kJ/kg\ K$ and the specific heat of air at constant volume is $0.718\ kJ/kg\ K$ .Find the specific gas constant.

  1. $0.287\ KJ/kg K$
  2. $0.21\ kJ/kg K$
  3. $0.34\ kJ/kg K$
  4. $0.19\ kJ/kg K$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Specific gas constant = Specific heat at constant pressure - Specific heat at constant volume

                                     = 1.005 - 0.718
                                     = 0.287 KJ/kgK

Multiple choice principal and molar specific heats of gases isothermal and adiabatic processes specific heat capacity heat and thermodynamics physics

The specific heat of Argon at constant volume is $0.3122 kj/kg K$. Find the specific heat of Argon at constant pressure if  $ R$  $=$8.314 kJ/Kmole K. (Molecular weight of argon$=$ $39.95$)

  1. $520.3$
  2. $530.2$
  3. $230.5$
  4. $302.5$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation
Given,
$C _v=0.3122\ kJ/kg.K$
$R=8.314$
$M=39.95$
$C _{p}=?$
We know,

$C _p-C _v=\dfrac{R}{M}$

$C _p-C _v=\dfrac{8.314}{39.95}=0.2081$

$C _p=0.3122+0.2081=0.5203$

$C _p=520.3\ J/kg.K$

Option $\textbf A$ is the correct answer
Multiple choice principal and molar specific heats of gases isothermal and adiabatic processes specific heat capacity heat and thermodynamics physics

Four moles of a perfect gas heated to increase its temperature by ${2^ \circ }C$ absorbs heat of 40 cal at constant volume. If the same gas is heated at constant pressure the amount of heat supplied is (R$=$ 2 cal/mol K)

  1. 28 cal

  2. 56 cal

  3. 84 cal

  4. 94 cal
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation
Heat supplied at constant volume
$Q _v=nC _v\triangle T$
$40=4\times C _v\times 2$
$C _v=5$ cal/mol.K
$C _p=C _v+R=5+2=7cal/mol.k$
$\Rightarrow $ Heat supplied at constant pressure
$Q _p=nC _p\triangle T=4\times 7\times 2$
$Q _p=56cal$
Multiple choice principal and molar specific heats of gases isothermal and adiabatic processes specific heat capacity heat and thermodynamics physics

The specific heat at constant volume for the monatomic argon is $0.075 \ kcal/kg-K$, whereas its gram molecular specific heat is $C _v \ = 2.98 \ cal/mol/K$. The mass of the argon atom is (Avogadro's number $= 6.02 \times 10^{23}$ molecules/mol)

  1. $6.60 \times 10^{-23} g$
  2. $3.30 \times 10^{-23}g$
  3. $2.20 \times 10^{-23}g$
  4. $13.20 \times 10^{-23}g$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Mass of one mole of argon =$\dfrac{gram \  molecular \  specific \  heat}{specific\   heat \  at \  constant \  volume}=\dfrac{2.98\times 10^{-3}}{0.075}=0.039733 \ g$


Thus mass of each argon atom=$\dfrac{0.0397333}{6.02\times 10^{23}}=6.60\times 10^{-23}g$

Multiple choice principal and molar specific heats of gases isothermal and adiabatic processes specific heat capacity heat and thermodynamics physics

If the ratio of specific heat of a gas at constant pressure to that at constant volume is $\gamma$, the change in internal energy of the mass of gas, when the volume changes from $V \ to \ 2V$ at constant pressure P, is

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

At constant pressure, change in internal energy$ \Delta U=nC _v\Delta T$
Now,$\dfrac{C _p}{C _v}=\gamma$
$\Rightarrow 1$+$\dfrac{R}{C _v}$=$\gamma$
$\Rightarrow C _v=\dfrac{R}{\gamma -1}$
Using Charle's law, final temperature=2\times initial temperatue=2T
Thus,  $ \Delta U=nC _v(2T-T)=nC _vT=\dfrac{nRT}{\gamma -1}=\dfrac{PV}{\gamma -1}$

Multiple choice principal and molar specific heats of gases isothermal and adiabatic processes specific heat capacity heat and thermodynamics physics

A vessel of volume $0.2 m^3$ contains hydrogen gas at temperature $300 K$ and pressure $1 \ bar$. Find the heat (in kcal) required to raise the temperature to $400 K$. (The molar heat capacity of hydrogen at constant volume is $5 \ cal/mol K$)

  1. $4$
  2. $2$
  3. $5$
  4. $8$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Using the equation $PV=nRT$ we have
$0.2\times 10^5=n\times 8.314\times 300$
Thus we get n as 8 moles.
Now the heat absorbed is given as 
$Q=W+U=nR\Delta T+nC _v\Delta T$
or
$Q=n(1+\frac{3}{2})R\Delta T$
or
$Q=8\times \frac{5}{2}\times 1.987\times 100=4000  kcal$

Multiple choice principal and molar specific heats of gases isothermal and adiabatic processes specific heat capacity heat and thermodynamics physics

The specific heat of a gas 

  1. Has only two value CP and Cv

  2. Has a unique value at a given temperature
  3. Can have any value between 0 and $\infty $

  4. Depends upon the mass of the gas
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

The specific heat of a gas is the amount of heat required to raise the temperature of unit mass by one degree, and depending on how heat is added (at constant pressure, constant volume, or polytropic processes), it can take any value between zero and infinity.

Multiple choice principal and molar specific heats of gases isothermal and adiabatic processes specific heat capacity heat and thermodynamics physics

A monatomic gas expands at constant pressure on heating. The percentage of heat supplied that increases the internal energy of the gas and that is involved in the expansion is

  1. 75%, 25%

  2. 25% 75%

  3. 60%, 40%

  4. 40%, 60%
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For isobaric expansion of monatomic gas,
Heat supplied, $\Delta Q=nC _p\Delta T=2.5nR\Delta T$,
Internal energy change, $\Delta U=nC _v\Delta T=1.5nR\Delta T$,
External work,$ \Delta W=P\Delta V=nR\Delta T$
So the heat energy is distributed as 3:2 between internal energy and work, i.e. 60%, and 40% respectively.

Multiple choice principal and molar specific heats of gases isothermal and adiabatic processes specific heat capacity heat and thermodynamics physics

The density of a polyatomic gas in standard conditions is $0.795 kg/m^3$. The specific heat of the gas at constant volume is

  1. $930\:J/kgK$
  2. $1400\:J/kgK$
  3. $1120\:J/kgK$
  4. $1600\:J/kgK$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For given polyatomic gas, applying ideal gas equation PM=dRT under standard conditions, where P=pressure=1atm, M=molar mass, d=density=$0.795kg/m^3$, 

R=universal gas constant, 
T=absolute temperature=$273K$

M=$0.795\times 8.31\times \dfrac{273}{100000}=0.018kg=18g$
Hence, molecule is $H _2O\Rightarrow$ degree of freedom =6
$\Rightarrow$ specific heat at constant volume=$C _v=\dfrac{f}{2}R=3R=3\times 8.31\times \dfrac{ 1000}{18}=1400\ J/kgK$

Multiple choice principal and molar specific heats of gases isothermal and adiabatic processes specific heat capacity heat and thermodynamics physics

A monatomic gas expands at constant pressure on heating. The percentage of heat supplied that increases the internal energy of the gas and that is involved in the expansion is

  1. $75\%$, $25\%$
  2. $25\%$, $75\%$
  3. $60\%$, $40\%$
  4. $40\%$, $60\%$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

According to the first law of thermodynamics,
Q = U + W
Now, at constant pressure, 
W = $ P \Delta V = nR \Delta T $
U = $ n {C} _{v} \Delta T $
For, a monoatomic gas, $ {C} _{v} = 1.5 R $
Thus, Q = $ 2.5 \ nR \Delta T $


Now, $ \dfrac{U}{Q} = \dfrac{1.5R}{2.5R} $ = 60 %
Similarly, $ \dfrac{W}{Q} = \dfrac{R}{2.5 R} $ = 40 %

Multiple choice principal and molar specific heats of gases isothermal and adiabatic processes specific heat capacity heat and thermodynamics physics

The value of $C _p-C _v=1.00:R$ for a gas in state $A$ and $C _p-C _v=1.06:R$ in another state. If $P _A$ and $P _B$ denote the pressure and $T _A$ and $T _B$ denote the temperatures in the two states, then

  1. $P _A=P _B$, $T _A>T _B$
  2. $P _A>P _B$, $T _A=T _B$
  3. $P _A < P _B$, $T _A>T _B$
  4. $P _A=P _B$, $T _A < T _B$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Since we know that,
$C _p-C _V=nR$
Therefore, state A contain less number of moles of gas then state B
Hence, Pressure in state A will be less than Pressure in state B 
whereas Temperature in state A will be greater than temperature in state B 
Since,
$P \propto n$
$T \propto 1/n$
Hence,
${ P } _{ A }<{ P } _{ B }$
$T _A>T _B$
option (C)