
The specific heat at constant volume for the monatomic argon is 0.075 kcal/kg – K, whereas its gram molecular specific heat is \[{C_V} = 2.98cal/molK\]. The mass of the argon atom is:
$\begin{array}{l}
{\rm{A}}{\rm{. 6}}{\rm{.60}} \times {\rm{1}}{{\rm{0}}^{ - 23}}g\\
{\rm{B}}{\rm{. 3}}{\rm{.30}} \times {\rm{1}}{{\rm{0}}^{ - 23}}g\\
{\rm{C}}{\rm{. 2}}{\rm{.20}} \times {\rm{1}}{{\rm{0}}^{ - 23}}g\\
{\rm{D}}{\rm{. 13}}{\rm{.20}} \times {\rm{1}}{{\rm{0}}^{ - 23}}g
\end{array}$
Answer
581.7k+ views
Hint: We are given the molar specific heat capacity and the gram molecular specific heat capacity for the argon gas from which we can calculate the molecular weight of one mole of the gas. One mole of a substance contains a number of atoms equal to the Avogadro number.
Formula used:
We have used the following formula for calculating molecular weight.
Gram molecular specific heat capacity $ = $ Molecular weight $ \times $ Molar specific heat capacity …(i)
1 mole $ = 6.02 \times {10^{23}}$atoms …(ii)
Complete step-by-step answer:
We are given the specific heat at constant volume for the monatomic gas which is basically the molar specific heat at constant volume. It is the specific heat capacity of the given monoatomic gas for 1 mole of the gas measured at constant volume. Its value is given to be
Molar specific heat$ = 0.075kcal/kg - K$
Then we are also given the gram molecular specific heat of the substance which is the specific heat capacity of the given monoatomic gas for 1 gram of the gas. The value of the gram molecular specific heat capacity of the given monatomic gas is given as
Gram molecular specific heat$ = 2.98cal/molK$
Now we can calculate the molecular weight of the argon gas by using the relation between the specific heats as given in equation (i). Using this equation, we can obtain the molecular weight of argon in the following way.
Molecular weight \[ = \dfrac{{{\rm{Gram molecular specific heat}}}}{{{\rm{Molar specific heat}}}}\]
From the given values, we get
Molecular weight \[ = \dfrac{{2.98}}{{0.075}} = 39.73g/mol\]
Now we can calculate the weight of one single atom of argon by substituting the value of 1 mole in the molecular weight which is the Avogadro’s number. It is done in the following way.
Mass of argon atom$ = \dfrac{{39.73}}{{6.02 \times {{10}^{23}}}} \simeq 6.60 \times {10^{ - 23}}g$
Hence the correct answer is option A.
Note: It is easier to understand the quantity which is being obtained in a calculation by checking the units of various quantities. Take care of the units while calculating the various quantities in this question.
Formula used:
We have used the following formula for calculating molecular weight.
Gram molecular specific heat capacity $ = $ Molecular weight $ \times $ Molar specific heat capacity …(i)
1 mole $ = 6.02 \times {10^{23}}$atoms …(ii)
Complete step-by-step answer:
We are given the specific heat at constant volume for the monatomic gas which is basically the molar specific heat at constant volume. It is the specific heat capacity of the given monoatomic gas for 1 mole of the gas measured at constant volume. Its value is given to be
Molar specific heat$ = 0.075kcal/kg - K$
Then we are also given the gram molecular specific heat of the substance which is the specific heat capacity of the given monoatomic gas for 1 gram of the gas. The value of the gram molecular specific heat capacity of the given monatomic gas is given as
Gram molecular specific heat$ = 2.98cal/molK$
Now we can calculate the molecular weight of the argon gas by using the relation between the specific heats as given in equation (i). Using this equation, we can obtain the molecular weight of argon in the following way.
Molecular weight \[ = \dfrac{{{\rm{Gram molecular specific heat}}}}{{{\rm{Molar specific heat}}}}\]
From the given values, we get
Molecular weight \[ = \dfrac{{2.98}}{{0.075}} = 39.73g/mol\]
Now we can calculate the weight of one single atom of argon by substituting the value of 1 mole in the molecular weight which is the Avogadro’s number. It is done in the following way.
Mass of argon atom$ = \dfrac{{39.73}}{{6.02 \times {{10}^{23}}}} \simeq 6.60 \times {10^{ - 23}}g$
Hence the correct answer is option A.
Note: It is easier to understand the quantity which is being obtained in a calculation by checking the units of various quantities. Take care of the units while calculating the various quantities in this question.
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