Class 11th

$\overrightarrow{L}=\overrightarrow{r}*\overrightarrow{p}\Rightarrow \overrightarrow{L}=mvr\left(\widehat{k}\right)$

Direction & magnitude both remain same

 for particle moving with constant speed.

$\frac{dP}{dV}=-aP,atV=V_0,P=P_0$

T_max =? for n = 1

$\frac{dP}{dV}=-aP$ (Given)

On integrating, we get

${\displaystyle \underset{P_0}{\overset{P}{\int }}\frac{dP}{P}}=-a{\displaystyle \underset{0}{\overset{v}{\int }}dV}$

$T_{max}=\frac{P_0}{aeR}?n=1$

$M{V}_0=M{V}_1+m{V}_2....\left(i\right)$

$\therefore M{V}_1=m{V}_2$  . (ii)

$M{V}_0^2=M{V}_1^2+m{V}_2^2$ . (iii)

$M{V}_0^2=M{V}_1^2+m{\left(\frac{M{V}_1}{m}\right)}^2\left[usinge{q}^n\left(ii\right)\right]$

$\left(1+\frac{M}{m}\right)=4$

${\left(\frac{M}{m}\right)}_{max}=3$

The main difference lies in the direction. We should know that in a longitudinal wave, the displacement of particles is parallel to the direction of wave propagation. In a transverse wave, however, the displacement is perpendicular to the direction of wave propagation.

The units of angular wave number and angular frequency are slightly different.

• Angular wave number k has units of radian per metre (rad/m).
• Angular frequency has units of radian per second (rad/s).

In Physics, we need to know that the phase angle tells us the position of a point within one cycle of a wave. That is measured in radians. It tells you how much a wave is ahead or, behind the other. On the other hand, the initial phase (or phase constant) is the phase angle at time t=0. It tells us where the wave begins with its oscillation cycle.

Applying : ${\left(n_1+n_2\right)}_{initial}={\left(n_1+n_2\right)}_{final}$  

Assuming the system attains a final temperature of T (Such that 300 < T < 60)

(Heat lost by N₂ of container l) = (Heat gained by N₂ of container II)

$n_{1}Cm\left(300-T\right)=n_{2}Cm\left(T-60\right)$  

$\left(\frac{2.8}{28}\right)\left(300-T\right)=\frac{0.2}{28}\left(T-60\right)$  

14 (300 – T) = T – 60

$P=\left(\frac{3}{28}\right)*8.31*\frac{284}{3*10^{-3}}*10^{-5}bar=0.84287bar$  

$P=84.28*10^{-2}bar$

$\approx 84*10^{-2}bar$  

$\kappa =\frac{1}{R}*\frac{\mathcal{l}}{A}=\left[\left(\frac{1}{1500}\right)*1.14\right]Sc{m}^{-1}=\frac{1.14}{1500}Sc{m}^{-1}$

${\lambda }_m=\frac{\kappa }{M}*1000Sc{m}^{2}mo{l}^{-1}$

${\lambda }_m=1000*\frac{\left(\frac{1.14}{1500}\right)}{0.001}Sc{m}^{2}mo{l}^{-1}=760Sc{m}^{2}mo{l}^{-1}$

Minimum temperature at which reaction becomes spontaneous is,

  $T_{min}=\frac{\Delta H^0}{\Delta S^0}$             

$\Delta {H^o}_{min}=\left[\Delta H_f^o\left(Fe\right)+\Delta H_f^o\left(CO\right)\right]-\left[\Delta H_f^o\left(FeO\right)+\Delta H_f^o\left(C\right)\right]$

$=\left[0-110.5\right]-\left[-266.3+0\right]$

$=155.8KJ/mol$

$\Delta S°=\left[\Delta S°\left(Fe\right)+\Delta S^o\left(CO\right)\right]-\left[\Delta S°\left(FeO\right)+\Delta S°\left(C\right)\right]$

$=\left(27.28+197.6\right)-\left(57.49+5.74\right)J/molK$

$=161.65J/molK$

$T_{min}=\frac{155.8*10^3}{161.65}K=963.8K\approx 964K$