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

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 \%$

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

If for hydrogen $C _p-C _v=m$ and for nitrogen $C _p-C _v=n$, where $C _p$ and $C _v$ refer to specific heats per unit mass respectively at constant pressure and constant volume, the relation between $m$ and $n$ is (molecular weight of hydrogen$=2$ and molecular weight of nitrogen$=14$)

  1. $n=14m$
  2. $n=7m$
  3. $m=7n$
  4. $m=14n$
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

For hydrogen, $C _P-C _V=\dfrac{1}{M _{H _2}}\dfrac{dQ}{dT}=m$

For nitrogen, $C _P-C _V=\dfrac{1}{M _{N _2}}\dfrac{dQ}{dT}=n$
$\implies \dfrac{m}{n}=\dfrac{M _{N _2}}{M _{H _2}}=\dfrac{14}{2}=7$

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

The average degree of freedom per molecule for a gas are $6$. The gas performs $25 J$ of work when it expands at a constant pressure. The heat absorbed by gas is 

  1. $75 \ J$
  2. $100 \ J$
  3. $150\ J$
  4. $125 \ J$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

For a gas with 'n' degrees of freedom:
$\gamma = 1 + \dfrac{2}{n} = 1 + \dfrac{2}{6} = \dfrac{4}{3}$
$C _{p} = \dfrac{\gamma R}{\gamma - 1} = 4R$
$C _{v} = \dfrac{R}{\gamma - 1} = 3R$

Heat supplied for constant pressure process is $nC _{p}\Delta T$

Change in internal energy $nC _{v} \Delta T$
$\dfrac{\Delta U}{Q} = \dfrac{C _{v}}{C _{p}} = \dfrac{1}{\gamma} = \dfrac{3}{4}$

Hence $\dfrac{W}{Q} = 1 - \dfrac{\Delta U}{Q} = \dfrac{1}{4}$
$\dfrac{W}{Q}=1-\dfrac{3}{4}=\dfrac{1}{4}$
$\implies Q = 100J$

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

What is the ratio of specific heats of constant pressure and constant volume for $NH _3$

  1. 1.33

  2. 1.44

  3. 1.28

  4. 1.67
Reveal answer Fill a bubble to check yourself
C Correct answer
Explanation

Ammonia (NH3) is a polyatomic gas. The ratio of specific heats (gamma) for polyatomic gases is typically lower than that of diatomic gases (1.4), and 1.28 is the standard value for NH3.

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

A reversible adiabatic path on a P- V diagram foran ideal gas passes through state A where P = 0.7$\times $ ${ 10 }^{ 2  }$ N/${ m }^{ -2 }$ and v=0.0049 $ { m }^{ 3  }$, The ratio of specific heat of the gas is 1.4 , The slop of patch at A is:

  1. $2.0 \times{ 10 }^{ 3\quad }{ Nm }^{ -5 }$
  2. $1.0 \times{ 10 }^{ 3\quad }{ Nm }^{ -8}$
  3. $-2.0\times{ 10 }^{ 7\quad }{ Nm }^{ -3 }$
  4. $-1.0\times{ 10 }^{ 3\quad }{ Nm }^{ -5 }$
Reveal answer Fill a bubble to check yourself
A Correct answer
Multiple choice principal and molar specific heats of gases isothermal and adiabatic processes specific heat capacity heat and thermodynamics physics

The value of the ratio ${C} _{p}/{C} _{v}$ for hydrogen is $1.67$ a $30K$ but decreases to $1.4$ at $300K$ as more degrees of freedom become active. During this rise in temperature (assume H2 as ideal gas),

  1. ${C} _{p}$ remains constant but ${C} _{v}$ increases
  2. ${C} _{p}$ decreases but ${C} _{v}$ increases
  3. Both ${C} _{p}$ and ${C} _{v}$ decreases by the same amount
  4. Both ${C} _{p}$ and ${C} _{v}$ increase by the same amount
Reveal answer Fill a bubble to check yourself
D Correct answer
Explanation

As temperature increases, rotational and vibrational degrees of freedom become active for hydrogen gas, increasing the internal energy and thus increasing both molar specific heats Cp and Cv by the same amount according to equipartition theorem.

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

A polyatomic gas with six degrees of freedom does $25\ J$ of work when it is expanded at constant pressure. The heat given to the gas is

  1. $100\ J$
  2. $150\ J$
  3. $200\ J$
  4. $250\ J$
Reveal answer Fill a bubble to check yourself
A Correct answer
Explanation

Degree of freedom, $f=6$
$\Rightarrow C _v=\dfrac{fR}{2}=3R$
$\Rightarrow C _p=C _v+R=4R$
Also work done $=\Delta W$=25J


Thus for isobaric process applying first law,
Heat given($\Delta Q$) $=$ internal energy change$(\Delta U)+\Delta W$
$\Rightarrow nC _p\Delta T=nC _v\Delta T+\Delta W$
$\Rightarrow 4nR\Delta T=3nR\Delta T +25J$
$\Rightarrow nR\Delta T=25J$
Hence, $ \Delta Q=4\times 25=100J$

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

A gas expands against a constant external pressure of  $2.00 atm, $ increasing its volume by $ 3.40 L.$   Simultaneously, the system absorbs  $400 J $ of heat from its surroundings. What is  $ \Delta E ,$  in joules, for this gas?

  1. $- 689$
  2. $-289$
  3. $+400$
  4. $+289$
Reveal answer Fill a bubble to check yourself
B Correct answer
Explanation

Using the first law of thermodynamics, delta E = Q - W. Work done by the gas is P * delta V = 2.00 atm * 3.40 L = 6.80 L*atm. Converting to Joules (1 L*atm = 101.3 J), W = 6.80 * 101.3 = 688.84 J. Thus, delta E = 400 J - 688.84 J = -288.84 J, which rounds to -289 J.

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

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

For any ideal gas, Cp - Cv = R (in molar terms). If 'a' and 'b' refer to specific heats (per unit mass), then Cp - Cv = R/M, where M is molar mass. For hydrogen, M=2; for nitrogen, M=28. Thus, a = R/2 and b = R/28. Therefore, a/b = 28/2 = 14, so a = 14b.