Calculate the number of atoms in 39.4 g gold. Molar mass of gold is 197g mol^-1 .

1

(a)

Molar mass of gold is 197 g mol^-1 , the number of atoms
= 6 x 10²³​
\(\therefore\) Number of atoms in 39.4 g
= \(\frac { 6.0\times { 10 }^{ 23 }\times 39.4 }{ 197 }\)
= 1.2 x 10²³

When air is pumped into a cycle tyre the volume and pressure of the air in the tyre both are increased." What about Boyle's law in this case?

1

(a)

When air is pumped, more molecules are pumped and Boyle's law is stated for situation where number of molecules constant.

The volume of a given mass of a gas at 27⁰C, 1 atm is 100 cc. What will be its volume at 327⁰C?

1

(a)

Keeping p constant, we have
\({ V }_{ 2 }\) = \(\frac { { V }_{ 1 }{ T }_{ 2 } }{ { T }_{ 1 } }\)
= \(\frac { 100\times 600 }{ 300 } \)
= 200cc

Two molecules of a gas have speeds of 9 \(\times\)10⁶ ms^-1 and 1 \(\times\) 10⁶ ms^-1, respectively. What is the root mean square speed of these molecules ?

1

(a)

\({ v }_{ rms }\) = \(\sqrt { \frac { { v }_{ 1\quad }^{ 2 }+{ { v }_{ 2 }^{ 2 } } }{ 2 } } \)
= \(\sqrt { \frac { (9\times { 10 }^{ 6 })^{ 2 }+(1\times { 10 }^{ 6 })^{ 2 } }{ 2 } }\)

= \(\sqrt { \frac { (81+1)\times { 10 }^{ 12 } }{ 2 } }\)
= \(\sqrt { 41 } \times { 10 }^{ 6 }ms^{ -1 }\)

Calculate the ratio of the mean free paths of the molecules of two gases having molecular diameters \(1\overset { 0 }{ A } \) and \(2\overset { 0 }{ A } \). The gases may be considered under identical conditions of temperature, pressure and volume.

1

(a)

As, we know, mean free path, 
\(\lambda \propto \frac { 1 }{ { d }^{ 2 } } \)
\(Given,\ { d }_{ 1 }=1\overset { 0 }{ A } \ and\ { d }_{ 2 }=2\overset { 0 }{ A } \Rightarrow { \lambda }_{ 1 }:{ \lambda }_{ 2 }=4:1\)