Question

A gas expands adiabatically at constant pressure, such that its temperature $T\alpha \dfrac{1}{{\sqrt V }}$ .The value of $\dfrac{{{C_P}}}{{{C_V}}}$ of the gas is:
$\left( A \right)1.30$
$\left( B \right)1.50$
$\left( C \right)1.47$
$\left( D \right)2.00$

Answer 1

With help of the problem statement a gas expands adiabatically at constant pressure. We know whenever there is a sudden expansion it is an adiabatic operation. In an adiabatic relation between p and V is $p{V^\gamma } = k$ ,with the help of this relation we can get the value of $\gamma$ .Now we know the formula for $\gamma$ using that we get the ratio of $\dfrac{{{C_P}}}{{{C_V}}}$ .

Complete Step By Step Answer:
When there is change in pressure and volume of gas also produce a change in temperature then it is called an adiabatic change. It is a kind of change in which no heat is allowed to enter into or escape from the gas, in other words, we can also say that there is no exchange of heat between the gas and its surroundings.
We know the formula for adiabatic expansion that is,
$p{V^\gamma } = k \ldots \ldots \left( 1 \right)$
Where,
Pressure = p
Volume = V
Constant = k
$\dfrac{{{C_P}}}{{{C_V}}} = \gamma \ldots \ldots \left( 2 \right)$
We know the gas equation as,
$pV = nRT$
Where,
Pressure = P
Volume = V
Amount of substance = n
Ideal gas constant = R
Temperature = T
Here let us assume amount of substance, n = 1
We will get,
$pV = RT$
Rearranging the above equation we will get,
$p = \dfrac{{RT}}{V} \ldots \ldots \left( 3 \right)$
Putting equation $\left( 3 \right)$ in place of equation $\left( 1 \right)$ we will get,
$\dfrac{{RT}}{V}{V^\gamma } = k$
Taking R to the other side we will get,
$\dfrac{T}{V}{V^\gamma } = \dfrac{k}{R}$
$\Rightarrow T{V^{\gamma - 1}} = \dfrac{k}{R}$
$\Rightarrow T{V^{\gamma - 1}} = {\text{constant}} \ldots \ldots \left( 4 \right)$
Here $\dfrac{k}{R} = {\text{constant}}$
But from the problem we know,
$T\alpha \dfrac{1}{{\sqrt V }}$
We can also write the above equation as,
$T = K\dfrac{1}{{\sqrt V }}$
Where K is a proportionality constant.
Rearranging the above equation we will get,
$T{V^{\dfrac{1}{2}}} = K$
$\Rightarrow T{V^{\dfrac{1}{2}}} = {\text{constant}} \ldots \ldots \left( 5 \right)$
Thus using equation $\left( 4 \right)$ and $\left( 5 \right)$ we will get,
Comparing both of them we will get,
$\gamma - 1 = \dfrac{1}{2}$
$\Rightarrow \gamma = \dfrac{1}{2} + 1$
$\Rightarrow \gamma = \dfrac{3}{2} \ldots \ldots \left( 6 \right)$
From equation $\left( 2 \right)$ we get,
$\dfrac{{{C_P}}}{{{C_V}}} = \gamma$
Putting the above equation in equation $\left( 6 \right)$ we will get,
$\dfrac{{{C_P}}}{{{C_V}}} = \dfrac{3}{2}$
$\Rightarrow \dfrac{{{C_P}}}{{{C_V}}} = 1.50$
Therefore the correct option is $\left( B \right)$ .

Note:
Keep in mind before solving this problem take an amount of substance that is n as one or else you will make a mistake. And there is no SI unit for the final value as we are taking the ratio of two same terms as both ${C_P}$ and ${C_V}$ denoted the specific heat.