
For a process, the relation between temperature and volume is \[T{V^3}\]= constant. If a monatomic gas follows this process, find molar specific heat for this process
A. $\dfrac{{7R}}{6}$
B. $\dfrac{R}{3}$
C. All of this
D. none of this
Answer
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Hint: The process is polytropic. In this process we see that $P{V^n}$ = constant. Ratio of specific heat is \[\gamma = \dfrac{{CP}}{{CV}}\] is a factor in adiabatic processes and it helps in determining speed in gases. This ratio for an ideal monatomic gas is \[\gamma = 1.66\] and for air it is\[{\text{ }}\gamma = 1.4\], which a diatomic gas usually is. Gamma is seen in many fluids equations which are generally related to pressure or temperature, and volume during a compression or expansion process.
Formula used :
\[c = \left( {\dfrac{R}{{\gamma - 1}}} \right) + \left( {\dfrac{R}{{1 - n}}} \right)\]
Where,
R is universal gas constant.
T is the temperature
P is the pressure
V is the volume
Gamma is an adiabatic exponent.
Complete step by step answer:
As we know that,
$PV = nRT$
$T{V^3}$= Constant
$\left( {\dfrac{{PV}}{{nR}}} \right) \times {V^3}$= Constant
$P{V^4}$=Constant
n is 4 in the above equation
we know that in monatomic gas is \[\gamma = \dfrac{{CP}}{{CV}}\]
in monoatomic gas \[\dfrac{3}{2}\]
$\gamma = \dfrac{{5\dfrac{R}{2}}}{{3\dfrac{R}{2}}}$
On simplification we get,
$\gamma = \dfrac{5}{3}$
Now we will substitute the values in the formula that we know
$c = \left( {\dfrac{{\dfrac{R}{1}}}{{\dfrac{5}{3} - 1}}} \right) + \left( {\dfrac{R}{{1 - 4}}} \right)$
On simplification we get,
$c = \dfrac{{7R}}{6}$
So option A is the correct option for the given question. Option A is $\dfrac{{7R}}{6}$ . So this is the correct option. Other options are invalid. Option D is wrong as we have one option correct. Option B is also incorrect as the value is different. Hence Option C is also not valid in this situation.
So, the correct answer is Option (A).
Note:
-We have to know that there are different kinds of gases like monoatomic diatomics and many more like that. There various formulae used to know more about these gases.
-We must remember that monatomic gases are gases composed of particles that have single atoms, for example helium or sodium vapour, and in this way different from polyatomic gases.
-Diatomic molecules are other molecules which are composed of only two atoms, they can be the same or different.
Formula used :
\[c = \left( {\dfrac{R}{{\gamma - 1}}} \right) + \left( {\dfrac{R}{{1 - n}}} \right)\]
Where,
R is universal gas constant.
T is the temperature
P is the pressure
V is the volume
Gamma is an adiabatic exponent.
Complete step by step answer:
As we know that,
$PV = nRT$
$T{V^3}$= Constant
$\left( {\dfrac{{PV}}{{nR}}} \right) \times {V^3}$= Constant
$P{V^4}$=Constant
n is 4 in the above equation
we know that in monatomic gas is \[\gamma = \dfrac{{CP}}{{CV}}\]
in monoatomic gas \[\dfrac{3}{2}\]
$\gamma = \dfrac{{5\dfrac{R}{2}}}{{3\dfrac{R}{2}}}$
On simplification we get,
$\gamma = \dfrac{5}{3}$
Now we will substitute the values in the formula that we know
$c = \left( {\dfrac{{\dfrac{R}{1}}}{{\dfrac{5}{3} - 1}}} \right) + \left( {\dfrac{R}{{1 - 4}}} \right)$
On simplification we get,
$c = \dfrac{{7R}}{6}$
So option A is the correct option for the given question. Option A is $\dfrac{{7R}}{6}$ . So this is the correct option. Other options are invalid. Option D is wrong as we have one option correct. Option B is also incorrect as the value is different. Hence Option C is also not valid in this situation.
So, the correct answer is Option (A).
Note:
-We have to know that there are different kinds of gases like monoatomic diatomics and many more like that. There various formulae used to know more about these gases.
-We must remember that monatomic gases are gases composed of particles that have single atoms, for example helium or sodium vapour, and in this way different from polyatomic gases.
-Diatomic molecules are other molecules which are composed of only two atoms, they can be the same or different.
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