Question

A thermally insulated vessel contains an ideal gas of molecular mass M and ratio of specific heats \[\Upsilon \] . it is moving with speed V and is suddenly brought to rest. Assuming no heat is lost to the surroundings, its temperature increases by:
A.$\dfrac{\left( \Upsilon -1 \right)}{2\left( \Upsilon +2 \right)\text{R}}\text{ M}{{\text{V}}^{2}}$
B.$\dfrac{\left( \Upsilon -1 \right)}{2\Upsilon \text{r}}\text{ M}{{\text{V}}^{2}}$
C.$\dfrac{\Upsilon \text{ M}{{\text{V}}^{2}}}{2\text{ R}}$
D.$\dfrac{\left( \Upsilon -1 \right)}{2\text{ R}}\text{M}{{\text{V}}^{2}}$

Answer 1

Hint: The work done during the thermodynamic process is equal to change in internal energy of the gas. By substituting the formulas for W and$\vartriangle \text{V}$ in the equation W =$\vartriangle \text{V}$, the value of$\vartriangle \text{T}$ can be found.

Complete answer:
We know that work during thermodynamic process is equal to change in internal energy of the gas, that is
          W and$\vartriangle \text{V}$….. (1)
Here $\text{W}=\dfrac{1}{2}\text{M}{{\text{V}}^{2}}$ …. (2)
Now, change in internal energy$\vartriangle \text{U}$ in terms of R and constant \[\Upsilon \] by the formula
          \[\vartriangle \text{U}=\dfrac{\text{R}}{\left( \Upsilon -1 \right)}\text{ }\vartriangle \text{T}\] …. (3)
Putting values of equation (2) and (3) in equation (1), we get
          $\dfrac{1}{2}\text{M}{{\text{V}}^{2}}=\dfrac{\text{R}}{\left( \Upsilon -1 \right)}\text{ }\vartriangle \text{T}$
Or $\vartriangle \text{T}=\dfrac{\left( \Upsilon -1 \right)}{\text{2R}}\text{M}{{\text{V}}^{2}}$

So, the correct option is (D).

Note:
We know that PV = nRT
Also, for1 mole of gas,
     $\text{U}={{\text{N}}_{0}}\text{f}\left( \dfrac{1}{2}\text{ KT} \right)$ .. .. (A)
Where ${{\text{N}}_{0}}$= Avogadro’s number
     K = Boltzmann’s constant
     T = Temperature
     f = degree of freedom
The relation between $\Upsilon =\dfrac{\left( \text{f}+2 \right)}{\text{f}}$
So,
$\begin{align}
  & \Upsilon -1=1+\dfrac{2}{\text{f}}-1 \\
 & \Upsilon -1=\dfrac{2}{\text{f}} \\
\end{align}$
Or $\text{f}=\dfrac{2}{\left( \Upsilon -1 \right)}$
Putting this value in equation (A), we get
     $\begin{align}
  & \text{U}={{\text{N}}_{0}}\left( \dfrac{2}{\Upsilon } \right)\left( \dfrac{1}{2}\text{ KT} \right) \\
 & \text{U}=\dfrac{{{\text{N}}_{0}}\text{KT}}{\Upsilon -1} \\
 & \text{ }=\dfrac{{{\text{N}}_{0}}\text{T}}{\left( \Upsilon -1 \right)}\left( \dfrac{\text{R}}{{{\text{N}}_{0}}} \right) \\
 & \text{U}=\dfrac{\text{RT}}{\left( \Upsilon -1 \right)} \\
\end{align}$
Or
$\vartriangle \text{V}=\dfrac{\text{R}}{\left( \Upsilon -1 \right)}\vartriangle \text{T}$
This is the derivation of internal energy in terms of R,$\Upsilon $and T.