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Question: A thermally insulated vessel contains an ideal gas of molecular mass M and ratio of specific heats \...

A thermally insulated vessel contains an ideal gas of molecular mass M and ratio of specific heats Υ\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.(Υ−1)2(Υ+2)R MV2\dfrac{\left( \Upsilon -1 \right)}{2\left( \Upsilon +2 \right)\text{R}}\text{ M}{{\text{V}}^{2}}
B.(Υ−1)2Υr MV2\dfrac{\left( \Upsilon -1 \right)}{2\Upsilon \text{r}}\text{ M}{{\text{V}}^{2}}
C.Υ MV22 R\dfrac{\Upsilon \text{ M}{{\text{V}}^{2}}}{2\text{ R}}
D.(Υ−1)2 RMV2\dfrac{\left( \Upsilon -1 \right)}{2\text{ R}}\text{M}{{\text{V}}^{2}}

Explanation

Solution

The work done during the thermodynamic process is equal to change in internal energy of the gas. By substituting the formulas for W and△V\vartriangle \text{V} in the equation W =△V\vartriangle \text{V}, the value of△T\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△V\vartriangle \text{V}….. (1)
Here W=12MV2\text{W}=\dfrac{1}{2}\text{M}{{\text{V}}^{2}} …. (2)
Now, change in internal energy△U\vartriangle \text{U} in terms of R and constant Υ\Upsilon by the formula
△U=R(Υ−1) △T\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
12MV2=R(Υ−1) △T\dfrac{1}{2}\text{M}{{\text{V}}^{2}}=\dfrac{\text{R}}{\left( \Upsilon -1 \right)}\text{ }\vartriangle \text{T}
Or △T=(Υ−1)2RMV2\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,
U=N0f(12 KT)\text{U}={{\text{N}}_{0}}\text{f}\left( \dfrac{1}{2}\text{ KT} \right) .. .. (A)
Where N0{{\text{N}}_{0}}= Avogadro’s number
K = Boltzmann’s constant
T = Temperature
f = degree of freedom
The relation between Υ=(f+2)f\Upsilon =\dfrac{\left( \text{f}+2 \right)}{\text{f}}
So,
Υ−1=1+2f−1 Υ−1=2f \begin{aligned} & \Upsilon -1=1+\dfrac{2}{\text{f}}-1 \\\ & \Upsilon -1=\dfrac{2}{\text{f}} \\\ \end{aligned}
Or f=2(Υ−1)\text{f}=\dfrac{2}{\left( \Upsilon -1 \right)}
Putting this value in equation (A), we get
U=N0(2Υ)(12 KT) U=N0KTΥ−1  =N0T(Υ−1)(RN0) U=RT(Υ−1) \begin{aligned} & \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{aligned}
Or
△V=R(Υ−1)△T\vartriangle \text{V}=\dfrac{\text{R}}{\left( \Upsilon -1 \right)}\vartriangle \text{T}
This is the derivation of internal energy in terms of R,Υ\Upsilon and T.