11.
The temperature of an ideal gas is increased from $$120\,K$$ to $$480\,K.$$ If at $$120\,K$$ the root-mean-square velocity of the gas molecules is $$v,$$ at $$480\,K$$ it becomes
According to law of equipartition of energy, the energy per degree of freedom is $$\frac{1}{2}kT.$$ For a polyatomic gas with $$n$$ degrees of freedom, the mean energy per molecule $$ = \frac{1}{2}nkT$$
According to boyle's law, at constant temperature.
$$P \propto \frac{1}{{\;V}}\,\,{\text{or}}\,\,{P_1}\;{V_1} = {P_2}\;{V_2}$$
14.
A non-linear triatomic gas is filled inside a vessel. If $$'\alpha '$$ fraction of moles dissociate into individual atoms, then average degree of freedom for the mixture is : (neglect vibrational degrees of freedom)
15.
The gases carbon-monoxide $$\left( {CO} \right)$$ and nitrogen at the same temperature have kinetic energies $${E_1}$$ and $${E_2}$$ respectively. Then
The gases carbon-monoxide $$\left( {CO} \right)$$ and nitrogen $$\left( {{N_2}} \right)$$ are diatomic, so both have equal kinetic energy $$\frac{5}{2}kT,{\text{i}}{\text{.e}}{\text{.}}\,{E_1} = {E_2}.$$
16.
The value of $${C_p} - {C_v}$$ is $$1.00R$$ for a gas sample in state $$A$$ and is $$1.06R$$ in state $$B.$$ Let $${p_A},{p_B}$$ denote the pressure and $${T_A},{T_B}$$ denote the temperature of the states $$A$$ and $$B$$ respectively. Then most likely
For ideal gas $${C_p} - {C_v} = R$$
If $${C_p} - {C_v} = 1.06\,R,$$
then gas will be real gas. Thus pressure is high and temperature is low for real gas.
17.
The pressure of a gas is raised from $${27^ \circ }C$$ to $${927^ \circ }C.$$ The root mean square speed
A
is $$\sqrt {\left( {\frac{{927}}{{27}}} \right)} $$ times the earlier value
$$RMS$$ speed is defined as the square root of the mean of the squares of the random velocities of the individual molecules of a gas. From Maxwellian distribution law, $$RMS$$ speed is given by $${c_{rms}} = \sqrt {\left( {\frac{{3kT}}{m}} \right)} $$
$$ \Rightarrow {c_{rms}} \propto \sqrt T $$
For two different cases i.e. at two different temperatures
$$\therefore \frac{{{{\left( {{c_{rms}}} \right)}_1}}}{{{{\left( {{c_{rms}}} \right)}_2}}} = \sqrt {\frac{{{T_1}}}{{{T_2}}}} $$
Here, $${T_1} = {27^ \circ }C = 300\,K$$
$$\eqalign{
& {T_2} = {927^ \circ }C = 1200\,K \cr
& \therefore \frac{{{{\left( {{c_{rms}}} \right)}_1}}}{{{{\left( {{c_{rms}}} \right)}_2}}} = \sqrt {\frac{{300}}{{1200}}} = \frac{1}{2} \cr
& \Rightarrow {\left( {{c_{rms}}} \right)_2} = 2{\left( {{c_{rms}}} \right)_1} \cr} $$
Hence, root mean square speed will be doubled.
18.
The equation of state for $$5g$$ of oxygen at a pressure $$p$$ and temperature $$T,$$ when occupying a volume $$V,$$ will be
Mean free path $${\lambda _m} = \frac{1}{{\sqrt 2 \pi {d^2}n}}$$
where $$d$$ = diameter of molecule and $$d = 2r$$
$$\therefore {\lambda _m} \propto \frac{1}{{{r^2}}}$$
20.
The average translational energy and the $$rms$$ speed of molecules in a sample of oxygen gas at $$300\,K$$ are $$6.21 \times {10^{ - 21}}J$$ and $$484\,m/s$$ respectively. The corresponding values at $$600\,K$$ are nearly (assuming ideal gas behavior)