Question

. The specific heat at constant volume for the monoatomic argon is 0.075 kcal/kg-K, whereas
its gram molecular specific heat Cv = 2.98 cal/mole/K. The mass of the argon atom is
(1)6.60x10-23 gm (2)3.30x10-23 gm (3) 2.20x10-23 gm (4)13.20x10-23 gm
(Avogadro's number=6.02x1023 molecules / mole)​

Answer 1

Answer:

• Mass (M) of Argon atom is 6.60 × 10⁻²³ Grams.

Given:

1. Specific Heat at Constant Volume = 0.075 Kcal/Kg - K
2. Gram Molecular Specific Heat = 2.98 cal/mole/K.

Explanation:

$\rule{300}{1.5}$

From Relation we know,

$\large\bigstar \: {\boxed{\tt C_v = M \times c_v}}$

$\bold{Here}\begin{cases}\tt{C_v} \: \text{Denotes Gram Molecular Specific Heat} \\ \text{at Constant Volume} \\ \\ \tt{c_v} \: \text{Denotes Specific Heat capacity} \\ \text{at Constant Volume}\end{cases}$

Now,

$\large{\boxed{\tt C_v = M \times c_v}}$

Substituting the values,

$\large{\tt \longmapsto 2.98 = M \times 0.075}$

But Specific Heat at Constant Volume is given in Kilo calories It needs to be Converted to Calories.

Therefore,

$\large{\tt \longmapsto 2.98 = M \times 0.075 \times 10^3}$

$\large{\tt \longmapsto 2.98 = M \times 75}$

$\large{\tt \longmapsto M = \dfrac{2.98}{75}}$

$\large{\tt \longmapsto M = \dfrac{298}{75 \times 100}}$

$\large{\tt \longmapsto M = \dfrac{1}{100} \times \dfrac{298}{75}}$

$\large{\tt \longmapsto M = \dfrac{1}{100} \times \cancel{\dfrac{298}{75}}}$

$\large{\tt \longmapsto M = \dfrac{1}{100} \times 3.97}$

$\large{\tt \longmapsto M = 3.97 \times 10^{-2}}$

$\large{\tt \longmapsto M = 39.7 \times 10^{-3}}$

$\large{\tt \longmapsto M = 39.7 \times 10^{-3} \: Kg}$

We Know,

$\large\star\: {\tt 1 \: gram \longrightarrow 10^{-3} \: Kg}$

Substituting,

$\large{\tt \longmapsto M = 39.7 \times 1 \: gram}$

$\large\longmapsto{\underline{\boxed{\tt M = 39.7 \: grams}}}$

This is the Mass of One Mole of atoms

$\rule{300}{1.5}$

$\rule{300}{1.5}$

Now, We Know, The Relation.

$\large\bigstar \: {\boxed{\tt m = \dfrac{M}{N_A}}}$

$\bold{Here}\begin{cases}\text{m Denotes Mass of the atom} \\ \text{M Denotes Mass of One Mole of atoms} \\ \tt{N_A} \text{ Denotes Avogadro's Number}\end{cases}$

Now,

$\large{\boxed{\tt m = \dfrac{M}{N_A}}}$

Substituting the values,

$\large{\tt \longmapsto m = \dfrac{39.7 \: grams}{6.02 \times 10^{23}}}$

$\large{\tt \longmapsto m = \dfrac{39.7}{6.02 \times 10^{23}}}$

$\large{\tt \longmapsto m = \dfrac{39.7}{6.02} \times 10^{-23}}$

$\large{\tt \longmapsto m = \cancel{\dfrac{39.7}{6.02}} \times 10^{-23}}$

$\large\longmapsto{\underline{\boxed{\red{\tt m = 6.60 \times 10^{-23} \: gm}}}}$

∴ Mass (M) of Argon atom is 6.60 × 10⁻²³ Grams (Option- 1).

$\rule{300}{1.5}$

Answer 2

Answer:

Option (1) 6.60 * 10-23 gm