Kinetic Theory Of Gases Question 6
Question: 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}=2.98,cal/mol/k $ . The mass of the argon atom is (Avogadro’s number $ =6.02\times 10^{23} $ molecules/mol)
Options:
A) $ 6.60\times {{10}^{-23}}g $
B) $ 3.30\times {{10}^{-23}}g $
C) $ 2.20\times {{10}^{-23}}g $
D) $ 13.20\times {{10}^{-23}}g $
Show Answer
Answer:
Correct Answer: A
Solution:
[a] Molar specific heat = molecular weight $ \times $ gram specific heat $ C _{v}=M\times c _{v} $
$ 2.98,cal/mol-K=M\times 0.075Kcal/kg-K $
$ =M\times \frac{0.075\times 10^{3}}{10^{3}}cal/g-K $
$ \
Therefore $ Molecular weight of argon $ M=\frac{2.98}{0.075}=39.7g $ i.e., mass of $ 6.023\times 10^{23} $ atom = 39.7 g
Therefore, mass of single atom $ =\frac{39.7}{6.023\times 10^{23}}=6.60\times {{10}^{-23}}g $
BETA
