Tag: heat and thermodynamics

Questions Related to heat and thermodynamics

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

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: C
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.

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

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

  1. $930:J/kgK$

  2. $1400:J/kgK$

  3. $1120:J/kgK$

  4. $1600:J/kgK$

Show answer
Correct Option: B
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=$C _v=\dfrac{f}{2}R=3R=3\times 8.31\times \dfrac{ 1000}{18}=1400\ J/kgK$

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

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: C
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 %

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

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

  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. $P _A=P _B$, $T _A < T _B$

Show answer
Correct Option: C
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)

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 $R=2:cal/mol^\circ C$)

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

  1. $750:cal$

  2. $630:cal$

  3. $1050:cal$

  4. $1470:cal$

Show answer
Correct Option: C
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 $
$Q = 1050 \ cal$

'n' number of liquids of masses m,2m,3m,4m, .......... having specific heats S, 2S, 3S, 4S, ...... at temperatures t, 2t, 3t, 4t, ........ are mixed. The resultant temperature of the mixture is

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

  1. $\frac{3n}{2n+1} t$

  2. $\frac{2n(n+1)}{3(2n+1)} t$

  3. $\frac{3n(n+1)}{2(2n+1)} t$

  4. $\frac{3n(n+1)}{(2n+1)} t$

Show answer
Correct Option: B

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

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

  1. $ \dfrac{1}{\gamma}$

  2. $\displaystyle(1- \dfrac{1}{\gamma})$

  3. $ \gamma -1$

  4. $\displaystyle(1- \dfrac{1}{\gamma^2})$

Show answer
Correct Option: B
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$=\dfrac{\gamma -1}{\gamma}=1-\dfrac{1}{\gamma}$

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

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

Show answer
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:

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

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