Tag: principal and molar specific heats of gases

Questions Related to principal and molar specific heats of gases

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)

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

  1. $6.60 \times 10^{-23} : g$

  2. $3.30 \times 10^{-23} : g$

  3. $2.20 \times 10^{-23} : g$

  4. $13.20 \times 10^{-23} : g$

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Correct Option: A
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.

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 $(R=2:cal/mol K)$.

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

  1. $40:kg$

  2. $40\times 10^{-3}:kg$

  3. $20:kg$

  4. $20\times 10^{-3}:kg$

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Correct Option: B
Explanation:

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

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

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   

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

  1. $1.77:kg/m^3$

  2. $0.77:kg/m^3$

  3. $1.77:g/m^3$

  4. $0.77:g/m^3$

Show answer
Correct Option: A
Explanation:

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

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

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 

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

  1. $75\%, 25\%$

  2. $25\%, 75\%$

  3. $60\%, 40\%$

  4. $40\%, 60\%$

Show answer
Correct Option: D
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 \%$

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$=14$)

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

  1. $n=14m$

  2. $n=7m$

  3. $m=7n$

  4. $m=14n$

Show answer
Correct Option: C
Explanation:

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

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

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 

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

  1. $75 \ J$

  2. $100 \ J$

  3. $150\ J$

  4. $125 \ J$

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Correct Option: B
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 $\dfrac{W}{Q} = 1 - \dfrac{\Delta U}{Q} = \dfrac{1}{4}$
$\dfrac{W}{Q}=1-\dfrac{3}{4}=\dfrac{1}{4}$
$\implies Q = 100J$

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

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

  1. 1.33

  2. 1.44

  3. 1.28

  4. 1.67

Show answer
Correct Option: C

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:

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

  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. $-1.0\times{ 10 }^{ 3\quad }{ Nm }^{ -5 }$

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Correct Option: A

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),

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

  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

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Correct Option: C

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

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

  1. $100\ J$

  2. $150\ J$

  3. $200\ J$

  4. $250\ J$

Show answer
Correct Option: A
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$