physics ncert solutions class 11th

$T=27°C,P=1atm=10^{5}N/m^2,V_{rms}=200m/s$

As we know

$v_{rms}=\sqrt[]{\frac{3RT}{M}}\Rightarrow \frac{v_{rms}}{v{\text{'}}_{rms}}=\frac{\sqrt[]{\frac{3RT}{M}}}{\sqrt[]{\frac{3RT\text{'}}{M}}}$

$V{\text{'}}_{rms}=200*\frac{2}{\sqrt[]{3}}=\frac{x}{\sqrt[]{3}}$

x = 400

[h] = ML²T^-1

[E] = ML²T^-2

[V] = ML²T^-2C^-1

[P] = MLT^-1

According to KTG, the gas exerts pressure because its molecule :

Suffer change in momentum when impinge on the walls of container.

According to question, we can write

Total moles of gas = n = n_Oxygen + n_Oxygen = $\frac{16}{2}+\frac{128}{32}=12moles$  

Volume of gas = 12 * 22.4 litre = 268.8 litre = 2.688 * 10⁵ cm³ $\approx 27*10^{4}c{m}^3$  

$\stackrel{?}{L}=\stackrel{?}{r}*m\stackrel{?}{v}$

= $\left(3\widehat{i}?\widehat{j}\right)*\left(3\widehat{j}+\widehat{k}\right)$

$=9\widehat{k}+3\left(?\widehat{j}\right)?\widehat{i}$

$=?\widehat{i}?3\widehat{j}+9\widehat{k}$

$\left|\stackrel{?}{L}\right|=\sqrt[]{1+9+81}=\sqrt[]{91}$

${\lambda }_e=\frac{h}{P_e}=\frac{h}{\sqrt[]{2m_e{K}_e}}\Rightarrow {\lambda }_e^2=\frac{h^2}{2m_e{k}_e}\Rightarrow K_e=\frac{h^2}{2m_e{\lambda }_e^2}$ - (i)

${\lambda }_P=\frac{h}{P_P}=\frac{hc}{K_P}\Rightarrow K_P=\frac{hc}{{\lambda }_P}$ - - (ii)

              K_e = K_P

_{$\Rightarrow {\lambda }_P\alpha {\lambda }_e^2$}

Use formula for M.I.

 According to question, we can write

Total moles of gas = n = n_Oxygen + n_Oxygen = $\frac{16}{2}+\frac{128}{32}=12moles$  

Volume of gas = 12 * 22.4 litre = 268.8 litre = 2.688 * 10⁵ cm³ $?27*10^{4}c{m}^3$  

$V_{rms}\alpha \sqrt[]{T}|T_i=127°C=400k.$

$V_{rm{s}^1}=2V_{rms},m=0.056kg=56g$

T_f = 4T_i = 4 * 400 = 1600 k

$Q=n{c}_v\Delta T=\left(\frac{m}{M}\right)C_v\Delta T$

= $\left(\frac{56}{28}\right)\frac{5}{2}R\Delta T$

= $2*\frac{5}{2}*2*1200$

Q = 12 * 10³ cal

= 12 k cal

$\lambda =\frac{v}{\sqrt[]{2}\pi d^{2}n{N}_a}=\frac{RT}{\sqrt[]{2}\pi d^{2}p{N}_a}$

= $\frac{\frac{25}{3}*300}{\sqrt[]{2}*\pi *10^{-20}*10^5*6*10^{23}}$

$x^3=\frac{v}{n{N}_a}$

$={\left(\frac{\frac{25}{3}*300}{10^5*6*10^{23}}\right)}^{1/3}$