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 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

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)

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

Five moles of hydrogen gas are heated from $30^\circ C$ to $60^\circ C$ at constant pressure. Heat given to the gas is (given $R=2:cal/mol^\circ C$)

  1. $750\:cal$
  2. $630\:cal$
  3. $1050\:cal$
  4. $1470\:cal$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

At constant pressure, the heat supplied to a gas is the same as its change in enthalpy,
Thus, Q = $ n {C} _{p} \Delta T $
$ {C} _{p} = 3.5 R $
Q = $ 5 \times 7 \times 30 $
$Q = 1050 \ cal$

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

The gas is heated at a constant pressure. The fraction of heat supplied used for external work is 

  1. $ \dfrac{1}{\gamma}$
  2. $\displaystyle(1- \dfrac{1}{\gamma})$
  3. $ \gamma -1$
  4. $\displaystyle(1- \dfrac{1}{\gamma^2})$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Heat absorbed =$ \Delta Q=nC _p\Delta T=\gamma nC _v\Delta T$
Internal energy change=$\Delta U=nC _v\Delta T$

Work done=$ \Delta Q- \Delta U=(\gamma -1)nC _v\Delta T$
Required fraction$=\dfrac{\gamma -1}{\gamma}=1-\dfrac{1}{\gamma}$

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 monoatomic argon is $0.075 : kcal/kg-K$, whereas its gram molecular specific heat is $C _v = 2.98 \ cal/molK$. The mass of the argon atom is (Avogrado'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 atoms=Gram molecular specific heat/Specific Heat

$=\dfrac{2.98g}{0.075 \ mol}=39.733\ g/mol$
Thus mass of one atom of argon=$\dfrac{39.733}{6.02\times 10^{23}}g$
$=6.60\times 10^{-23}g$
Hence correct answer is option A.

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

The mass of a gas molecule can be computed from the specific heat at constant volume. $C _v$ for argon is $0.075:kcal/kg K$. The molecular weight of an argon atom is $(R=2:cal/mol K)$.

  1. $40\:kg$
  2. $40\times 10^{-3}\:kg$
  3. $20\:kg$
  4. $20\times 10^{-3}\:kg$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

$C _v=75\ cal/kgK=75\times M _o cal/molK=\dfrac{3}{2}R$

$\implies 75M _o=\dfrac{3}{2}\times2kg$
$\implies M _o=0.04kg=40\times 10^{-3}kg$

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

The specific heats of argon at constant pressure and constant volume are $525:J/Kg$ and $315:J/Kg$, respectively. Its density at NTP will be   

  1. $1.77\:kg/m^3$
  2. $0.77\:kg/m^3$
  3. $1.77\:g/m^3$
  4. $0.77\:g/m^3$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

$m(C _P-C _V)T=PV$

$\implies C _P-C _V=\dfrac{PV}{mT}=\dfrac{P}{dT}$
$\implies d=\dfrac{P}{(C _P-C _V)T}=\dfrac{1.01\times 10^{5}}{273\times (525-315)}=1.77\ kg/m^3$

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

A monoatomic gas expands at a 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
D 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 \%$