Principal and molar specific heats of gases - class-XI

$N _2$ is a diatomic molecule thus its degree of freedom is 5. Its $C _p$ is given as $(1+\displaystyle\dfrac{f}{2})R=(1+\dfrac{5}{2})R=\dfrac{7}{2}R$
Thus, we get the heat required as $Q=nC _p\Delta T=0.2\times \displaystyle\dfrac{7}{2}\times 8.314\times 300=1746  J$

When the gas is heated at constant pressure, some amount of heat is used up in increasing the volume of the gas. For a constant volume process no such heat is required. Thus $C _p>C _v$

We know that $\displaystyle\dfrac{C _p}{C _v}=\gamma$ and $C _p-C _v=R$.
Thus we get $C _v(\displaystyle\dfrac{C _p}{C _v}-1)=R$
or, $C _v=\displaystyle\dfrac{R}{\gamma -1}$

Helium is monatomic (gamma = 1.67) and Oxygen is diatomic (gamma = 1.4). Using the formula for the mixture ratio of specific heats, the result is approximately 1.62.

Using the relationship Cp - Cv = R/M and Cp/Cv = gamma, we can solve for Cv using the density and pressure at NTP. The calculation yields approximately 2.162 kcal/kg K.

The correct answer is option(B).

The different formula for specific heats is given by:

• $\dfrac{C _{p}}{C _{v}} = \gamma$
• $C _{p} - C _{v} = R$

• $C _{p} = \dfrac{\gamma R}{\gamma -1}$
• $C _{v} = \dfrac{R}{\gamma -1}$

From given data we have $C _{p} - C _{v} = 2 $
and $\dfrac{C _{p}} { C _{v}} = \dfrac {5}{3} $
Solving both gives , 
$C _{p} = 5$ and  $ C _{v} = 3 $

The diatomic molecule has total 6 degress of freedom(3 translational, 2 rotational and 1 vibrational)
Now $C _p$ is given as $(1+\dfrac{f}{2})R=(1+\dfrac{6}{2})R=4R$
and $C _v$ is given as $\dfrac{f}{2}R=\dfrac{6}{2}R=3R$
Thus we get $\dfrac{C _p}{C _v}=\dfrac{4}{3}=1.33$

C$ _p$-C$ _v$ = R and hence option D is incorrect

The change in internal energy in the process should have been,
U = $ nC _v \Delta T $
Now, for this process, if $ \dfrac{{C} _{p}}{{C} _{v}} = \gamma $ and $C _p-C _v=R$
Then, $C _v =  \dfrac{R}{\gamma - 1} $
U = $ \dfrac{nR \Delta T}{\gamma - 1} $
Now, $ nR \Delta T = P(2V - V) $
Thus, U = $ \dfrac{PV}{\gamma - 1} $

In an isobaric process, pressure is constant.
Heat added in an isobaric process is given by
$\Delta Q=\Delta U+ \Delta W$
$nC _p\Delta T=nC _v\Delta T+nR\Delta T$
$\displaystyle \therefore \dfrac {\Delta Q}{\Delta W}=\dfrac {nC _p\Delta T}{nR\Delta T}=\dfrac {C _p}{R}=\dfrac {\gamma R/\gamma -1}{R}=\dfrac {\gamma }{\gamma -1}$
Option B.

Since, the process is at constant volume,
Q = U as W = 0
Thus, Q = $ n {C} _{v} \Delta T $
At STP, n = $ \dfrac{PV}{RT} $
Since, He is diatomic, $ {C} _{v} = 2.5R $
Q = $ \dfrac{PV}{RT} \times 2.5R \times 15 $
Substituting the pressure and temperature values at STP, 
P = 1 atm
V = 67.2 L
T = 298 K
we get,
Q = 560.9 J

In a diatomic gas, we have $C _p=\dfrac{7}{2}R $ and $C _v=\dfrac{5}{2}R$
The heat is given as $nC _p\Delta T$ and internal energy as $nC _v\Delta T$
Thus we get $\dfrac{U}{Q}$ as $\dfrac{5}{7}$

C$ _{v}$ cannot be greater than $C _{p}$

By definition 

$dU={ C } _{ v }dT\longrightarrow 1$

also enthalpy,

$H=U+PV\\ or\quad dH=dU+d\left( PV \right) \\ or\quad dH=dU+nRdT\longrightarrow 2$

Also $dH={ C } _{ P }dT\\ \therefore { C } _{ P }dT={ C } _{ V }dT+nRdT\\ \Rightarrow { C } _{ P }={ C } _{ V }+nR\\ or{ C } _{ P }-{ C } _{ V }\quad =nR=\cfrac { Rm }{ M } $

for $m=1$

${ C } _{ P }-{ C } _{ V }=\cfrac { R }{ M } $

For ideal gas
${C} _{P}-{C} _{V}=R/M$
If ${C} _{P}$ and ${C} _{V}$ are specific heats $\left( J/kg- _{  }^{ o }{ C } \right) $
$M=$ molar mass of gas
$\Rightarrow a=R/2$ and $b=R/28$
$\Rightarrow$ $a=14b$

For any ideal gas,$C _p-C _v=nR$, where $R$ is the gas constant.
That is $C _p-C _v$ per mole for any gas is a constant value.
So, $a=b$
Option D.

The final temperature of a mixture of gases is the weighted average of the temperatures, where the weights are the number of molecules (or moles) of each gas.

According to Mayer's relation, the difference between molar specific heats at constant pressure and constant volume is equal to the universal gas constant R for any ideal gas, meaning Cp - Cv = R for both hydrogen and oxygen. Thus, a equals b.

Specific gas constant = Specific heat at constant pressure - Specific heat at constant volume

Mass of one mole of argon =$\dfrac{gram \  molecular \  specific \  heat}{specific\   heat \  at \  constant \  volume}=\dfrac{2.98\times 10^{-3}}{0.075}=0.039733 \ g$

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}$

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
$Q=8\times \frac{5}{2}\times 1.987\times 100=4000  kcal$

The specific heat of a gas is the amount of heat required to raise the temperature of unit mass by one degree, and depending on how heat is added (at constant pressure, constant volume, or polytropic processes), it can take any value between zero and infinity.

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.

For given polyatomic gas, applying ideal gas equation PM=dRT under standard conditions, where P=pressure=1atm, M=molar mass, d=density=$0.795kg/m^3$, 

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 $

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)

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 $
$Q = 1050 \ cal$

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

Mass of one mole of argon atoms=Gram molecular specific heat/Specific Heat

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

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

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 %$

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

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

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.

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.

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

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 Latm. Converting to Joules (1 Latm = 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.

The room volume is 150 m^3. Using PV=nRT, calculate the number of moles of air. The total heat added by 50 people is 50 * 150 W = 7500 J/s. The heat required to raise the air temperature from 20 C to 37 C is Q = n * Cv * delta T. Solving for time t = Q / Power gives approximately 4.34 minutes.

Cp is greater than Cv because at constant pressure, the system does work on the surroundings as it expands, requiring additional heat input to achieve the same temperature rise. The reason correctly explains the assertion.

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.

According to Mayer's relation $C _{p}-C _{v}= R/m$

From Mayer's relation Cp - Cv = R and the definition of adiabatic index gamma = Cp / Cv, we can substitute Cp = gamma * Cv into Mayer's relation to get Cv(gamma - 1) = R, which rearranges to Cv = R / (gamma - 1).

Using the equipartition theorem, internal energy U = (f/2)RT, so Cv = dU/dT = (f/2)R. Then Cp = Cv + R = ((f/2) + 1)R. Taking their ratio gamma = Cp / Cv gives 1 + (2 / f).

For an ideal gas, the ratio of molar specific heats is given by gamma = Cp / Cv. Using the relation Cp - Cv = R, we find Cv = Cp - R = (7/2)R - R = (5/2)R. Therefore, gamma = (7/2)R / ((5/2)R) = 7/5.

$\gamma =\dfrac{C _p}{C _v}$ i.e, ratio of specific heat capacity at constant pressure and specific heat capacity at constant volume. It doesn't depend on temperature, i.e, it is independent of temperature.

For all ideal gases, the difference between molar heat capacity at constant pressure and constant volume is defined by the universal gas constant, Cp - Cv = R. This value is independent of the atomicity of the gas.

 For an ideal gas, $C _p$ is greater than $C _v$ because when gas is heated at constant volume, whole of the heat supplied is used to increase the temperature only but when gas is heated at constant pressure, the heat supplied is used to increases both temperature and the volume of gas (heat is used to do work)

The adiabatic exponent, often denoted by gamma, is defined as the ratio of the molar heat capacity at constant pressure to the molar heat capacity at constant volume.

Using the first law dQ = dU + dW, where dQ = C dT, dU = Cv dT, and dW = P dV. Substituting C = Cv + betaV, we get (Cv + betaV) dT = Cv dT + P dV. This simplifies to betaV dT = P dV. Using P = RT/V, we get betaV dT = (RT/V) dV, which leads to dT/T = (R/beta) * (dV/V^2). Integrating this leads to the process equation.

Total heat capacity at constant volume (Cv_total) is defined as Q / delta T. Given Q = 80 J and delta T = 20 K, Cv_total = 80 / 20 = 4.0 J/K.

Heat supplied to one mole of gas in a constant pressure process is given by: $Q = C _{p}\Delta T$
Change in the internal energy of gas is given by:$\Delta U = C _{v}\Delta T$

As it is a constant pressure process, using Charles law we get $\displaystyle\frac{V}{T}=constant$. Thus for a threefold increase in volume we get threefold increase in temperature. Thus we get the final temperature as $3(273)=819 K.$ Thus $\Delta T=819-273=546 K$.
Now as helium is a diatomic molecule, its degree of freedom f is 5. Ths we get $C _p$ for it as $(1+\displaystyle\frac{f}{2})R=(1+\frac{5}{2})R=\frac{7}{2}R$.
Thus heat transferred will be given as $\Delta Q=nC _p\Delta T$
or
$\Delta Q=1(\displaystyle\frac{7}{2})(1.987)(546)=3820\  cal$

For a diatomic gas we have the degree of freedom as 5. 

$C _v$ is given as $\displaystyle\dfrac{f}{2}R$. Here $f$ is the degree of freedom. For monoatomic gas $f=3$ and for diatomic gas $f=5$. 

$C _p$ for a gas is given as $(1+\displaystyle\dfrac{f}{2})R$ and $C _v$ is given as $\displaystyle\dfrac{f}{2}R$. Here f is the degree of freedom. 

Duplicate of 492697. NH3 is nonlinear polyatomic with γ = 1.28 at room temperature, accounting for partial vibrational contribution.

The heat supplied at constant pressure is given by Q = n * Cp * Delta T. Substituting the given values: 300 = 2 * (5R/2) * Delta T, where R = 8.314 J/mol K. Solving for Delta T gives 300 / (5 * 8.314) = 300 / 41.57 = 7.22 degrees Celsius.

A real gas behaves most like an ideal gas under conditions of low pressure and high temperature. Under these conditions, intermolecular forces and the actual volume of gas molecules become negligible compared to empty space and thermal energy.

Given P proportional to T^2 and P V = n R T, substituting P gives T^2 V proportional to T, meaning V proportional to 1/T or V T = constant. The molar heat capacity C is given by Cv + R / (1 - x) where V T^x = constant here x = 1, leading to C = Cv + R / (1 - 1) which needs differentiation properly. Using Q = Delta U + W, where Delta U = Cv dT and W = P dV, with PV = nRT and P = k T^2, we find C = R/2 for a monatomic gas.

I is correct (heat added). II is correct (work done by system). III is incorrect because if W=0, no work is done by the system; the system is at constant volume.

$Q = nC _p \Delta T$ and $W = P\Delta V = nR\Delta T$
monatomic gas, $\displaystyle{C _p = \dfrac{5R}{2}}$.
$\Rightarrow$$\displaystyle{\dfrac{W}{Q} = \dfrac{2}{5}}$

For isobaric process.
$ \Delta Q= n C _{p} \Delta T$
$=2 \times \dfrac{5}{2} R \times (35-30)$
$= 208 \ J$

$dU = nC _v dT$ or, $ 80 = 5 \times C _v(120 - 100)$
$C _v = 4.0\ J/K$

Let $Cu$ and $Cp$ be molar specific heats of the ideal gas at a 

For a diatomic gas, the specific heat at constant pressure $C _p=\frac{7}{2}R$ and the specific heat at constant volume $C _v=\dfrac{5}{2}R$

Hellium is monoatomic.

Given $\gamma =5/3$

Heat added for a constant pressure process is,

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

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

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 $13841.97\ J/Kg-k$

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

We know that for one mole of gas,

$Th\quad fraction\quad is\quad \frac { \triangle V }{ \triangle Q } \quad =\quad \frac { { _{ n }{ C } _{ v } }\triangle T }{ { _{ n }{ C } _{ p } }\triangle T } \ \frac { \triangle V }{ \triangle Q } =\frac { { C } _{ V } }{ { C } _{ P } } \quad =\quad \frac { 1 }{ Y } \ as\quad we\quad know\quad y\quad =\quad { C } _{ P }/{ C } _{ V }\ y\quad for\quad diatomatic\quad gas\quad :\quad \ { C } _{ P }\quad of\quad diatometic\quad gas\quad :\quad \frac { 7 }{ 2 } \ { C } _{ V }\quad of\quad diatometic\quad gas\quad :\quad \frac { 5 }{ 2 } \ y\quad =\quad \frac { { C } _{ P } }{ { C } _{ V } } =\frac { 7/2 }{ 5/2 } =\frac { 7 }{ 5 } \ \frac { \triangle V }{ \triangle Q } =\frac { 1 }{ y } =\frac { 1 }{ 7/5 } =\frac { 5 }{ 7 } \quad (D)$

When  heat  is
 supplied  at  constant  volume,  temperature  increases accordingly  to  the  ideal  gas  equation.

$P=\dfrac { nRT }{ V } $

as  V  is  constant  and  T  is  increasing,  pressure  will  also  increase.

Than at constant pressure  as temperature is increase volume increases, resulting in expansion of the gas, resulting in positive work, Hence the heat given is used up for expansion and then to increases the internal energy . The heat capacity at constant pressure is larger.