
One mole of an ideal monoatomic gas is mixed with $1$ mole of an ideal diatomic gas. The molar specific heat of the mixture at constant volume is:
A. $3cal$
B. $4cal$
C. $8cal$
D. $9cal$
Answer
435.3k+ views
Hint: We have to see, the molar warmth limit of a synthetic substance is the measure of energy that should be added, as warmth, to one mole of the substance to cause an expansion of one unit in its temperature. On the other hand, it is the warmth limit of an example of the substance partitioned by the measure of substance of the example; or likewise the particular warmth limit of the substance times its molar mass.
Complete answer:
We have to know that the temperature of an example of a substance mirrors the normal active energy of its constituent particles comparative with its focal point of mass. Quantum mechanics predicts that, at room temperature and common pressing factors, a disconnected molecule in a gas cannot store any critical measure of energy besides as dynamic energy. In this way, when a specific number $N$ of particles of a monatomic gas gets an info $\Delta Q$ of warmth energy, in a compartment of fixed volume, the dynamic energy of every iota will increment by $\dfrac{{\Delta Q}}{N}$ , autonomously of the molecule's mass. This supposition is the establishment of the hypothesis of ideal gases.
At the end of the day, that hypothesis predicts that the molar warmth limit at steady volume ${C_{V,m}}$ of all monatomic gases will be something similar; explicitly,
${C_{V,m}} = \dfrac{3}{2}R$
A similar hypothesis predicts that the molar warmth limit of a monatomic gas at consistent pressing factor will be
${C_{P,m}} = {C_{V,m}} + R = \dfrac{5}{2}R$
To calculate the average of both the monoatomic and diatomic values,
${C_V} = \dfrac{{\left[ {\dfrac{3}{2}R + \dfrac{5}{2}R} \right]}}{2} = 2R = \left( {2 \times 2} \right) = 4cal$
Hence, the correct option is (B).
Note:
We have to see in particular the warmth, estimated the molar warmth limit of a substance, particularly a gas, might be fundamentally higher when the example is permitted to extend as it is warmed than when it is warmed in a shut vessel that forestalls extension (at constant volume). The proportion between the two, notwithstanding, is a similar warmth limit proportion got from the relating explicit warmth limits.
Complete answer:
We have to know that the temperature of an example of a substance mirrors the normal active energy of its constituent particles comparative with its focal point of mass. Quantum mechanics predicts that, at room temperature and common pressing factors, a disconnected molecule in a gas cannot store any critical measure of energy besides as dynamic energy. In this way, when a specific number $N$ of particles of a monatomic gas gets an info $\Delta Q$ of warmth energy, in a compartment of fixed volume, the dynamic energy of every iota will increment by $\dfrac{{\Delta Q}}{N}$ , autonomously of the molecule's mass. This supposition is the establishment of the hypothesis of ideal gases.
At the end of the day, that hypothesis predicts that the molar warmth limit at steady volume ${C_{V,m}}$ of all monatomic gases will be something similar; explicitly,
${C_{V,m}} = \dfrac{3}{2}R$
A similar hypothesis predicts that the molar warmth limit of a monatomic gas at consistent pressing factor will be
${C_{P,m}} = {C_{V,m}} + R = \dfrac{5}{2}R$
To calculate the average of both the monoatomic and diatomic values,
${C_V} = \dfrac{{\left[ {\dfrac{3}{2}R + \dfrac{5}{2}R} \right]}}{2} = 2R = \left( {2 \times 2} \right) = 4cal$
Hence, the correct option is (B).
Note:
We have to see in particular the warmth, estimated the molar warmth limit of a substance, particularly a gas, might be fundamentally higher when the example is permitted to extend as it is warmed than when it is warmed in a shut vessel that forestalls extension (at constant volume). The proportion between the two, notwithstanding, is a similar warmth limit proportion got from the relating explicit warmth limits.
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