Practicing Success

Target Exam

CUET

Subject

Physics

Chapter

Thermodynamics

Question:

The specific heat at constant volume for the monoatomic argon is 0.075 kcal/kg-K, whereas its gram molecular specific heat is Cv = 29.8 cal/mol-K. The mass of the argon atom is : (Avogadro's number = 6.02*10^{23}\) molecules/mol : 

Options:

6.60 x 10-23 g

3.30 x 10-23 g

2.20 x 10-23 g

13.20 x 10-23 g

Correct Answer:

6.60 x 10-23 g

Explanation:

Molar specific heat = molecular weight x gram specific heat

2.98 cal/mol-K = M x 0.075 kcal/kg-K

  = M x \(\frac{0.075*10^3}{10^3}\) cal/g-K

\(\Rightarrow \text{ Molecular weight of argon : M } = \frac{2.98}{0.075}\) = 39.7 g

i.e. mass of 6.023 x 1023 atom = 39.7 g

Therefore, mass of single atom = \(\frac{39.7}{6.023*10^{23}}\) = 6.60 x 10-23 g