Tag: isothermal and adiabatic processes

Questions Related to isothermal and adiabatic processes

A gaseous mixture consists of $16\ g$ of helium and $16\ g$ of oxygen, then the ratio $\dfrac { { C } _{ p } }{ { C } _{ v } } $of the mixture is

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

  1. $1.4$

  2. $1.54$

  3. $1.59$

  4. $1.62$

Show answer
Correct Option: D
Explanation:

Hellium is monoatomic.

Oxygen is diatomic.
Degree of freedom of He, ${ f } _{ 1 }=$3
Degree of freedom of ${ O } _{ 2 }$, ${ f } _{ 2 }=$5
$\therefore \cfrac { { C } _{ p } }{ { C } _{ v } }$ of mixture.
$\cfrac { { C } _{ p } }{ { C } _{ v } } =\cfrac { [{ N } _{ 1 }(2+{ f } _{ 1 })+{ N } _{ 2 }(2+{ f } _{ 2 })] }{ { N } _{ 1 }{ f } _{ 1 }+{ N } _{ 2 }{ f } _{ 2 } }$
Putting,${ N } _{ 1 }={ N } _{ 2 }=16\quad gm,\quad$ and $\ { f } _{ 1 }=3,\quad { f } _{ 2 }=5$
we get,$ \cfrac { { C } _{ p } }{ { C } _{ v } } =\cfrac { 3 }{ 2 } \ =1.62$

When a heat of Q is supplied to one mole of a monatomic gas $\left ( \gamma =5/3 \right )$, the molar heat capacity 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. $ \dfrac{3R}{4}$

  2. $ \dfrac{5R}{4}$

  3. $ \dfrac{7R}{4}$

  4. $\dfrac{3R}{2}$

Show answer
Correct Option: D
Explanation:

Given $\gamma =5/3$

i.e $\displaystyle \dfrac {C _p}{C _v}=\dfrac {5}{3}$
and $C _p-C _v=R$

$\therefore \displaystyle \dfrac {C _v+R}{C _v}=\dfrac {5}{3}$
$\displaystyle 1+\dfrac {R}{C _v}=\dfrac {5}{3}$

$C _v= \dfrac{3R}{2}$
Option D.

The molar specific heat of helium at constant volume is $3\ cal/mol^{o}C$ . Heat energy required to raise the temperature of 1gm helium gas by $1^{o}C$ at constant pressure is :

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

  1. 1.2 cal

  2. 1.25 cal

  3. 3 cal

  4. 4 cal

Show answer
Correct Option: B
Explanation:

Heat added for a constant pressure process is,


$dQ=dU+dW$

$nC _p\Delta T=nC _v\Delta T+nR\Delta T$

Given 1 gm of Helium, number of moles$= 1/4 =0.25$

$R=2\ cal/mol-K$

$dQ=0.25[3(1)+2(1)]=0.25(5)=1.25\ cal$

Option B.

When 5 moles of gas is heated from $100^{o}C$ to $120^{o}C$ at constant volume, the change in internal energy is 200 J. The specific heat capacity of the gas is

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

  1. $5\space Jmol^{-1}K^{-1}$

  2. $4\space Jmol^{-1}K^{-1}$

  3. $2\space Jmol^{-1}K^{-1}$

  4. $1\space Jmol^{-1}K^{-1}$

Show answer
Correct Option: C
Explanation:

Change in internal energy is given by
$dU=nC _v\Delta T$
$200=5(C _v)20$
$C _v=2J/mol.K$
Option C.

A mass of $50$ g of a certain metal at $150^0C$ is immersed in $100$ g of water at $11^0C.$ The final temperature is $20^0C$. Calculate the specific heat capacity of the metal. Assume that the specific heat capacity of water is $4.2 J g^{-1}K^{-1}$.

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

  1. $0.682 J g^{-1}K^{-1}$

  2. $582 J g^{-1}K^{-1}$

  3. $0.582 J g^{-1}K^{-1}$

  4. $0.0582 J g^{-1}K^{-1}$

Show answer
Correct Option: C
Explanation:

Mass if the solid , $m _s=50 g$

Initial temperature of the solid, $t _s=150^0 C=150+273=423 k$
Mass of water, $m _w=100 g$
Temperature of the water, $t _w=11^0 C=11+273=293 k$
According to  principle of calorie meter,
Heat gained by water =Heat lost by the solid 
$\begin{array}{l} \therefore { m _{ w } }{ C _{ w } }\left( { t-{ t _{ w } } } \right) ={ m _{ s } }{ C _{ s } }\left( { { t _{ s } }-t } \right)  \ \Rightarrow 100\times 4.2\left( { 293-284 } \right) =50\times { C _{ s } }\times \left( { 423-293 } \right)  \ \Rightarrow 3780=6500\, { C _{ s } } \ \therefore { C _{ s } }=0.582\, \, J/g\, \, k \end{array}$
Hence, Option $C$ is correct.

$n _{1}$ and $n _{2}$ moles of two ideal gases of the thermodynamics constant $\gamma _{1}$ and $\gamma _{2}$ respectively are mixed. $C _{p}/ C _{v}$ for the mixture is

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

  1. $\dfrac {\gamma _{1} + \gamma _{2}}{2}$

  2. $\dfrac {n _{1}\gamma _{1} + n _{2}\gamma _{2}}{n _{1} + n _{2}}$

  3. $\dfrac {n _{1}\gamma _{2} + n _{2}\gamma _{1}}{n _{1} + n _{2}}$

  4. $\dfrac {n _{1}\gamma _{1}(\gamma _{2} + 1) + n _{2}\gamma _{2}(\gamma _{1} - 1)}{n _{1}(\gamma _{1} - 1) + n _{2}(\gamma _{1} - 1}$

Show answer
Correct Option: D

A sphere of density $\rho$, specific heat capacity c and radius r, is hung by a thermally insulated thread in an enclosure which is kept at a temperature slightly lower than that of the sphere. The rate of change of temperature for the sphere depends upon the temperature difference between the sphere and the enclosure, and is proportional to then

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

  1. $\dfrac{c}{r^3 \rho}$

  2. $\dfrac{r^3 \rho}{c}$

  3. $r^3 \rho c$

  4. $\dfrac{1}{r \rho c}$

Show answer
Correct Option: D
Explanation:

$P. \dfrac{4}{3} \pi r^3 . c. \dfrac{dT}{dt} = e . 4 \pi r^2 \sigma (T - T _0)$
$\dfrac{dT}{dt} \propto \dfrac{1}{Prc}$

1g of $H _{2}$ gas is heated by $1^{o}C$ at constant pressure. The amount of heat spent in expansion of gas is

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

  1. $\dfrac{4.155}{4.18}cal$

  2. $\dfrac{4.7}{2.1}cal$

  3. $\dfrac{6.8}{2.2}cal$

  4. $\dfrac{1.26}{1.7}cal$

Show answer
Correct Option: A
Explanation:

The amount of heat spent in the process will be,
Q = $ nR \Delta T $
Q = $ \dfrac{1}{2} \times 8.314 \times 1 $ J
The same value in calorie will be, Q = $ \dfrac{4.155}{4.185} $ cal

The volume of $1\ kg$ of hydrogen gas at $N.T.P$ is $11.2\ m^{3}$. Specific heat of hydrogen at constant volume is $10046J\ kg^{-1}K^{-1}$. Find the specific heat at constant pressure.

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

  1. $13.8\ kJ/kg-K$

  2. $14.2\ kJ/kg-K$

  3. $16.4\ kJ/kg-K$

  4. $18.3\ kJ/kg-K$

Show answer
Correct Option: A
Explanation:

Given that,

Mass of hydrogen $m=1\,kg$

Volume of hydrogen $V=11.2\,{{m}^{3}}$

Specific heat of hydrogen at constant volume ${{C} _{V}}=10046\,JK{{g}^{-1}}{{k}^{-1}}$

 We know that,

N.T.P condition as follows

  $ P=1.01\times {{10}^{5}}\,N/{{m}^{2}} $

 $ T={{25}^{0}}C=298\,K $

 Applying gas equation

 $PV=nRT$

 Putting the values in the above equation

 $ PV=nRT $

$ 1.01\times {{10}^{5}}\times 11.2=1\times R\times 298 $

$ R=3795.97\,J/kgk $

 Now according to the Mayer's law

 ${{C} _{P}}-{{C} _{V}}=R$

 Putting the values in the above equation

 $ {{C} _{P}}-{{C} _{V}}=R $

 $ {{C} _{P}}=3795.97+10046 $

$ {{C} _{P}}=13841.97\,JK{{g}^{-1}}{{k}^{-1}} $

Hence, the specific heat of hydrogen at constant pressure is $13841.97\ J/Kg-k$

Molar heat capacity of an ideal gas whose molar heat capacity at constant is $C _v$ for process $P=2e^{2v}$( where P is pressure of gas and V is volume of gas)

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

  1. $C _v + \dfrac{R}{1+2V}$

  2. $C _v + \dfrac{R}{2V}$

  3. $C _v + \dfrac{R}{V}$

  4. None of these

Show answer
Correct Option: A
Explanation:

$\begin{array}{l} By\, u\sin  g\, \, first\, law\, of\, Ther{ { moodynamics } }:- \ dQ=dw+dU \ and,\, also\,  \ dQ=nCdT \ dw=Pdv \ dU=n{ C _{ v } }dT \ Now,\, substituting\, them\, in\, the\, first\, law\, we\, get \ \Rightarrow nCdT=PdV+n{ C _{ n } }dT \ C=\frac { { PdV } }{ { ndT } } +{ C _{ v } } \ To\, find\, \, \frac { { PdV } }{ { ndT } } \, we\, will\, use\, the\, ideal\, gas\, equation \ PV=nRT \ 2V{ e^{ 2V } }=nRT\, \, \, \, \, \, \left[ { \, { { Re } }place\, \, P=2{ e^{ 2v } } } \right]  \ Differentiating\, both\, sides\, with\, respect\, to\, T \ 2\left( { { e^{ 2V } }+2V{ e^{ 2V } } } \right) \frac { { dV } }{ { dT } } =nR \ Now,\, from\, this\, we\, have \ \frac { { PdV } }{ { ndT } } =\frac { R }{ { 1+2v } }  \ So,\, we\, get \ C={ C _{ v } }+\frac { R }{ { 1+2v } }  \ Hence,\, the\, option\, A\, is\, the\, correct\, answer. \end{array}$