Question

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

Answer 1

Hint: An attempt to this question can be made by understanding the first law of thermodynamics. Use the formula derived from the first law of thermodynamics to arrive at the answer. The formula is given below:
$\Delta U = Q - W$
Where,
U is internal energy of the system
Q is the quantity of energy supplied to the system in the form of heat
W is the work done by the system on the surroundings


Complete Solution :The first law of thermodynamics is often considered as the law of conservation of energy adapted mainly for thermodynamic processes.
- It distinguishes heat from thermodynamic work and also relates them to a function of a body's condition called internal energy.
- The law of conservation of energy states that the total energy of an isolated system is constant. This means that energy can be transformed from one form to another. Energy is neither created nor destroyed.
In the question it is given to us that the heat transfer is 0.
Internal energy at constant volume is equal to $n{{C}_{V}}\Delta T$.

Since the body is in translation motion, the work done is calculated in accordance to the work energy theorem.
$W=-\dfrac{1}{2}M{{v}^{2}}$
Substituting the above values in the first law of thermodynamics:
$\Delta U=Q - W$
$0=\Delta U + W$
$n{{C}_{V}}\Delta T-\dfrac{1}{2}M{{v}^{2}}=0$
$\Delta T=\dfrac{\dfrac{1}{2}M{{v}^{2}}}{n{{C}_{V}}}$
$\Delta T=\dfrac{\dfrac{1}{2}M{{v}^{2}}}{\left( \dfrac{nR}{\gamma -1} \right)}$
Upon simplification, the change is temperature becomes, $\dfrac{\left( \gamma -1 \right)}{2R}M{{v}^{2}}K$
So, the correct answer is “Option C”.

Note: It is important to know that the first law of thermodynamics is applicable only for ideal gases. The equation used above ceases to exist in the case of real gases.