For 1 mole with f degrees of freedom,
Internal energy, U = 1 x C_v x T = f²/RT
For n moles, U = nC_vT = nf²/RT

Degree of freedom, f = 3N - K.

As force of attraction between molecules disappears, then the molecules will hit the wall with more speeds, hence , F = \(\frac{\Delta p}{\Delta t}\) , where F is average force on the wall due to the molecules.
\(\Delta\) p is change in momentum and \( \Delta \)t is the time duration . Due to increase in \(\Delta\) p, force F will also increase, hence pressure, p = \(\frac{F}{A}\) will increase. Here, A is area of one wall.

The average KE will be samea conditions of temperatue and pressure are same.
v_rms \(\propto\)  \(\frac{1}{\sqrt{m}}\)
\(\because\) m_A > m_B  > m_C
\(\Rightarrow\) v_C > v_B > v_A

Yes, the molar specific heat of a solid, a constant quantity as its value is 3R J/mol^-k.

Degree of freedom will increase when temperature is very high because at high temperature, vibrational motion of the gas will contribute to the kinetic energy. Hence, there is an additional kinetic energy associated with the gas, as a result of increased degree of freedom.

If a gas is heated, then the root mean square velocity of its molecules is increased.
∴ V_rms ∝ √T
∴ The temperature of the gas increases .

68.25 K

390 m/s

1.8 x 10³ ms^-1

\(As\ we\ know\ that,\ { V }_{ rms }=\sqrt { \frac { 3RT }{ M } } \)
\(\\ Thus,\ { V }_{ rms }/Ar={ V }_{ rms }/He\)
\(\\ \sqrt { \frac { { T }_{ Ar } }{ { M }_{ Ar } } } =\sqrt { \frac { { T }_{ He } }{ { M }_{ He } } } \)
\(\\ { T }_{ Ar }=?\ { T }_{ He }=273-10=263K\)
\(\\ { M }_{ Ar }=39.9u,{ M }_{ He }=4u\)
\(\\ Thus,\quad \frac { { T }_{ Ar } }{ 39.9 } =\frac { 263 }{ 4 } \)
\(\\ { T }_{ Ar }=\frac { 263\times 39.9 }{ 4 } =2623.43K\)

\(Given,\ Diameter\ molecule,d=2\overset { 0 }{ A }=2\times { 10 }^{ -10 }m\)
\(\\ At\quad STP,\ one\ mole\ of\ gas\ (or\ 22.4\ L)\ of\ gas\ have\)
\(\\ { N }_{ A }=6.023\times { 10 }^{ 23 }molecules\)
\(\\ \therefore Number\ density\ of\ nitrogen\ molecules\)
\(\\ n=\frac { { N }_{ A } }{ 22.4\quad L } =\frac { 6.023\times { 10 }^{ 23 }{ m }^{ -3 } }{ 22.4\times { 10 }^{ -3 }{ m }^{ 3 } } =2.69\times { 10 }^{ 25 }{ m }^{ -3 }\)
\(\\ \therefore Mean\ free\ path\ of\ nitrogen\ at\ STP\ condition,\)
\(\\ \lambda =\frac { 1 }{ \sqrt { 2\pi n{ d }^{ 2 }\ } } \)
\(\\ \lambda =\frac { 1 }{ 1.414\times 3.142\times (2.69\times { 10 }^{ 25 })\times (2\times { 10 }^{ -10 })^{ 2 } }\)
\( \\ =2.1\times { 10 }^{ -7 }m\)

\(For\ a\ monoatomic\ gas,\ i.e.\gamma =\frac { 5 }{ 3 }\)
\( \\ { C }_{ { V }_{ \gamma } }=\frac { R }{ \gamma -1 } =\frac { R }{ \frac { 5 }{ 3 } -1 } =\frac { 3 }{ 2 } R\)
\(\\ For\ a\ diamotic\ gas,\ i.e.\ \gamma =\frac { 7 }{ 5 }\)
\( \\ { C }_{ V }=\frac { R }{ \frac { 7 }{ 5 } -1 } =\frac { 5 }{ 2 } R\)
\(\\ By\ conservation\ of\ energy,\)
\(\\ { C }_{ { V }_{ mixture } }=\frac { { \mu }_{ 1 }{ C }_{ { V }_{ 1 } }+{ \mu }_{ 2 }{ C }_{ { V }_{ 2 } } }{ { \mu }_{ 1 }+{ \mu }_{ 2 } } \)
\(=\frac { 2\times \frac { 3 }{ 2 } R+3\times \frac { 5 }{ 2 } R }{ 2+3 } =\frac { 3R+7.5R }{ 5 } =2.1\ R\)

A given mass of water in vapour state has 1.67×10³ times the volume of the same mass of water in liquid state : For a given mass of water molecules, the density is less if volume is large. So the volume of the vapour is 1000/0.6 = 1/(6 × 10 ^-4 ) times larger. If densities of bulk water and water molecules are same, then the fraction of molecular volume to the total volume in liquid state is 1. As volume in vapour state has increased, the fractional volume is less by the same amount, i.e. 6×10^-4 .
This is also the increase in the amount of volume available for each molecule of water. When volume increases by 103 times the radius increases by V1/3 or 10 times, i.e., 10 × 2 Å = 20 Å. So the average distance is 2 × 20 = 40 Å.

In the liquid (or solid) phase, the molecules of water are quite closely packed. The density of water molecule may therefore, be regarded as roughly equal to the density of bulk water = 1000 kg m^–3. To estimate the volume of a water molecule, we need to know the mass of a single water molecule. We know that 1 mole of water has a mass approximately equal to
(2 + 16)g = 18 g = 0.018 kg.
Since 1 mole contains about 6 × 10²³ molecules (Avogadro’s number), the mass of a molecule of water is (0.018)/(6 × 10²³) kg = 3 × 10^–26 kg.
Therefore, a rough estimate of the volume of a water molecule is as follows :
Volume of a water molecule = (3 × 10^–26 kg)/ (1000 kg m^–3)
= 3 × 10^–29 m³ = (4/3) π (Radius)³
Hence, Radius ≈ 2 ×10^-10 m = 2 Å

6.21 x 10^-21 J

1.8 x 10³ K

\( V=250 \mathrm{cc}=250 \times 10^{-6} \mathrm{~m}^3 \)
\( \mathrm{P}=10^{-3 \mathrm{~mm}}=10^{-3} \times 10^{-3} \mathrm{~m} \)
\( =\left(10^{-6} \times 13600 \times 10\right) \)
\(=136 \times 10^{-3} \text { Pascal } \)
\( \mathrm{T}=27^0 \mathrm{C}=300 \mathrm{k} \)
\( \mathrm{n}=\frac{\mathrm{PV}}{R T} \)
\( =\frac{136 \times 10^{-3} \times 250 \times 10^{-6}}{8.3 \times 300}=1.36 \times 10^{-8}\)
No. of molecules
\(=1.36 \times 10^{-8} \times 6 \times 10^{23}  \)
\(=8.17 \times 10^{15}\)

Initially in the oxygen cylinder, V₁=30 litre =\(30 \times 10^{-3} \mathrm{~m}^3\),
\(P_1=15 a t m=15 \times 1.01 \times 10^5 \mathrm{~Pa}, T_1=27+273 =300 \mathrm{~K}\)
If the cylinder contains n₁ mole of oxygen gas, then \(P_1 V_1=n_1 R T_1\)
or
\(n_1=\frac{P_1 V_1}{R T_1}=\frac{\left(15 \times 1.01 \times 10^5\right) \times\left(30 \times 10^{-3}\right)}{8.3 \times 300} =18.253\)
For oxygen moleculaer weight, M=32 g
Initial mass of oxygen in the cylinder cylinder,
\(m_1=n_1 M=18.253 \times 32=548.1 g\)
Finally in the oxygen gas in the cylinder, let n₂ moles of oxygen be left,
Here,
\( V_2=30 \times 10^{-3} \mathrm{~m}^3, P_2=11 \times 1.01 \times 10^5 \mathrm{~Pa}, T_2 =17+273=290 \mathrm{~K}\)
Now,
\( n_2=\frac{P_2 V_2}{R T_2}=\frac{\left(11 \times 1.01 \times 10^5\right) \times\left(30 \times 10^{-3}\right)}{8.3 \times 290} =13.847 \)
\( \therefore \text { Final mass of oxygen gas in the cylinder, } m_2=13.847 \times 32=453.1 \mathrm{~g} \)
\( \therefore \text { Mass of the oxygen gas withdrawn }=m_1-m_2=584.1-453.1=131.0 \mathrm{~g} . \)