NEET-UG 2012

According to the question,

Damping force, F $\propto$ v

$\Rightarrow$ F = kv
where k is constant of proportionality.

$\therefore$ k = $\frac{F}{v}$ = $\frac{N}{m{s}^{-1}}$ = $\frac{kgm{s}^{-2}}{m{s}^{-1}}$ = $kg{s}^{-1}$

Given x = 8 + 12t $-$ t³

Velocity, v = $\frac{dx}{dt}$ = 12 - 3t²

When v = 0, then 12 - 3t² = 0

$\Rightarrow$ t = 2 s

$a=\frac{dv}{dt}$ = - 6t

$\therefore$ At t = 2 s, $a$ = - 12 m/s²

$\therefore$ Retardation = 12 m/s²

$\overrightarrow{P}$ = vector sum = $\overrightarrow{A}+\overrightarrow{B}$

$\overrightarrow{Q}$ = Vector differences = $\overrightarrow{A}-\overrightarrow{B}$

Since $\overrightarrow{P}$ and $\overrightarrow{Q}$ are perpendicular

$\therefore$ $\overrightarrow{P}.\overrightarrow{Q}=0\Rightarrow \left(\overrightarrow{A}+\overrightarrow{B}\right).\left(\overrightarrow{A}-\overrightarrow{B}\right)=0$

$\Rightarrow A^2=B^2=\left|\overrightarrow{A}\right|=\left|\overrightarrow{B}\right|$

Horizontal range

$R=\frac{u^2\sin 2\theta }{g}$     ......(1)

Maximum height

$H=\frac{u^2{\sin }^2\theta }{2g}$       .....(2)

According to the problem R = H

$\frac{u^2\sin 2\theta }{g}$ = $\frac{u^2{\sin }^2\theta }{2g}$

$\Rightarrow 2\sin \theta \cos \theta =\frac{{\sin }^2\theta }{2}$

$2\cos \theta =\frac{\sin \theta }{2}$

$\Rightarrow \cot \theta =\frac{1}{4}$

$\Rightarrow \tan \theta =4$

$\Rightarrow \theta ={\tan }^{-1}(4)$

At maximum compression the solid cylinder will stop so loss in K.E. of cylinder = gain in P.E. of spring

$\Rightarrow \frac{1}{2}m{v}^2+\frac{1}{2}I{\omega }^2=\frac{1}{2}k{x}^2$

$\Rightarrow \frac{1}{2}m{v}^2+\frac{1}{2}\frac{m{R}^2}{2}{\left(\frac{v}{R}\right)}^2=\frac{1}{2}k{x}^2$

$\Rightarrow \frac{3}{4}m{v}^2=\frac{1}{2}k{x}^2$

$\Rightarrow \frac{3}{4}\times 3\times {\left(4\right)}^2=\frac{1}{2}\times 200x^2$

$\Rightarrow \frac{36}{100}=x^2\Rightarrow x=0.6m$

Here, $U=\frac{A}{r^2}-\frac{B}{r}$

For equilibrium, $\frac{dU}{dr}=0$

$\therefore -\frac{2A}{r^3}+\frac{B}{r^2}=0$

$\Rightarrow \frac{2A}{r^3}=\frac{B}{r^2}\Rightarrow r=\frac{2A}{B}$

For stable equilibrium, $\frac{d^{2}U}{d{r}^2}>0$

$\frac{d^{2}U}{d{r}^2}=\frac{6A}{r^4}-\frac{2B}{r^3}$

${\frac{d^{2}U}{d{r}^2}|}_{r=\left(2A/B\right)}=\frac{6A{B}^4}{16A^4}-\frac{2B^4}{8A^3}=\frac{B^4}{8A^3}>0$

So for stable equilibrium, the distance of the particle is $\frac{2A}{B}$.

As no external force acts on the system, therefore centre of mass will not shift.

conservation of linear momentum along x-direction

$m_{2}v=m_1{v}_x\Rightarrow \frac{m_{2}v}{m_1}=v_x$

along y-direction

$m_2\times \frac{v}{2}=m_1{v}_y\Rightarrow v_y=\frac{m_{2}v}{2m_1}$

Note: Let A moves in the direction, which makes an angle q with initial direction i.e.

$\tan \theta =\frac{v_y}{v_x}=\frac{m_{2}v}{2m_1}/\frac{m_{2}v}{2m_1}$

$\tan \theta =\frac{1}{2}$

$\tan \theta =\frac{1}{2}\Rightarrow \theta ={\tan }^{-1}\left(\frac{1}{2}\right)$ to the x-axis.

Here, m = 1000 kg, R = 90 m, $\theta$ = 45°

For banking, $\tan \theta =\frac{v^2}{Rg}$

$\Rightarrow v=\sqrt{Rg\tan \theta }=\sqrt{90\times 10\times \tan {45}^o}=30m{s}^{-1}$

When a mass is rotating in a plane about a fixed point its angular momentum is directed along a line perpendicular to the plane of rotation.

Let x be the distance of centre O of equilateral triangle from each side. Total torque about O = 0

$\Rightarrow$ F₁x + F₂x – F₃x = 0

or F₃ = F₁ + F₂

Gravitational attraction force on particle B,

$F_g=\frac{G{M}_{P}m}{{\left(D_P/2\right)}^2}$

Acceleration of particle due to gravity

$a=\frac{F_g}{m}=\frac{4G{M}_P}{D_P^2}$

Let at h height, the weight of a body becomes 1/16th of its weight on the surface.

$W_h=16W_s$   ...(i)

$\Rightarrow$ $16W_s=W_s$   ...(ii)

$W_h=mg=\frac{GMm}{{\left(R_e+h\right)}^2}$

$g^{\prime }=\frac{GM}{{\left(R_e+h\right)}^2}$

Similarly, $g=\frac{GM}{R_e^2}$

$\frac{g^{\prime }}{g}=\frac{R_e^2}{{\left(R_e+h\right)}^2}\Rightarrow g^{\prime }=g{\left(1+\frac{h}{R}\right)}^{-2}$

$\Rightarrow W_h=16W_s$

$\Rightarrow g^{\prime }=\frac{g}{16}$

$\frac{1}{16}={\left(1+\frac{h}{R}\right)}^{-2}\Rightarrow 4=1+\frac{h}{R}$

h = 3R

According to Kelpner’s law of period T² $\propto$ R³

$\frac{T_1^2}{T_2^2}=\frac{R_1^3}{R_2^3}=\frac{{\left(6R\right)}^3}{{\left(3R\right)}^3}=8$

$\frac{24\times 24}{T_2^2}=8$

$T_2^2=\frac{24\times 24}{8}=72=36\times 2$

$\therefore$ $T_2=6\sqrt{2}$

Temperature of liquid oxygen will first increase in the same phase. Then, the liquid oxygen will change to gaseous phase during which temperature will remain constant, After that temperature of oxygen in gaseous state will increase. Hence option (a) represents corresponding temperature-time graph.

Stefan’s law for black body radiation

$Q=\sigma eA{T}^4$

$\therefore T={\left[\frac{Q}{\sigma \left(4\pi R^2\right)}\right]}^{1/4}$

In a cyclic process,

$\Delta U=0$

In a cyclic process work done is equal to the area under the cycle and is positive if the cycle is clockwise and negative if anticlockwise.

$\therefore \Delta$W = – Area of rectangle ABCD = – P(2V) = – 2PV

According to first law of thermodynamics

$\Delta Q=\Delta U+\Delta W\Rightarrow \Delta Q=\Delta W$
    (As $\Delta U=0$)

i.e., heat supplied to the system is equal to the work done

So heat absorbed, $\Delta$Q = $\Delta$W = – 2PV

$\therefore$ Heat rejected by the gas = 2PV

From the above P-V diagrams, (d) is correct, as from the question, the initial gas goes from volume V to 3V and then volume of the gas get reduced from 3V to V at constant pressure. In case of an isothermal expansion, P-V curve is rectangular hyperbola which is stated by (d).

Let $l$ be the length of the string. Fundamental frequency is given by

$\upsilon =\frac{1}{2l}\sqrt{\frac{T}{\mu }}$

$\Rightarrow v\propto \frac{1}{l}$  ($\because$ T and $\mu$ are constants)

Here, $l_1=\frac{k}{{\upsilon }_1},l_2=\frac{k}{{\upsilon }_2},l_3=\frac{k}{{\upsilon }_3}$ and $l=\frac{k}{\upsilon }$

But $l=l_1+l_2+l_3$

$\frac{1}{\upsilon }=\frac{1}{{\upsilon }_1}+\frac{1}{{\upsilon }_2}+\frac{1}{{\upsilon }_3}$

$2\pi f_1=600\pi$

$f_1=300$   ...(i)

$2\pi f_2=608\pi$

$f_2=304$   ...(ii)

$\left|f_1-f_2\right|=4beats$

$\frac{I_{\max }}{I_{m.n}}=\frac{{\left(A_1+A_2\right)}^2}{{\left(A_1+A_2\right)}^2}=\frac{{\left(5+4\right)}^2}{{\left(5-4\right)}^2}=\frac{81}{1}$

where A₁, A₂ are amplitudes of given two sound wave.

Let the side length of square be 'a' then potential at centre O is



$V=\frac{k(-Q)}{\frac{a}{\sqrt{2}}}+\frac{k(-q)}{\frac{a}{\sqrt{2}}}+\frac{k(2q)}{\frac{a}{\sqrt{2}}}+\frac{k(2Q)}{\frac{a}{\sqrt{2}}}=0$

= – Q – q + 2q + 2Q = 0 = Q + q = 0

$\therefore$ Q = – q

Eight identical cubes are required so that the given charge q appears at the centre of the bigger cube.
Thus, the electric flux passing through the given cube is

$\varphi =\frac{1}{8}\left(\frac{q}{{\epsilon }_0}\right)=\frac{q}{8{\epsilon }_0}$

Torque, $\tau$ = pEsin$\theta$

Potential energy, U = –pEcos$\theta$

Since the galvanometer shows no deflection so current will flow as shown in the figure.


Current,

$I=\frac{V_A}{R_1+R}=\frac{12V}{\left(500+100\right)\Omega }=\frac{12}{600}A$

$V_B=IR=\left(\frac{12}{600}A\right)\left(100\Omega \right)=2V$

Resistance of bulb is constant

$P=\frac{V^2}{R}\Rightarrow \frac{\Delta p}{p}=\frac{2\Delta V}{V}+\frac{\Delta R}{R}$

$\therefore \frac{\Delta p}{p}=2\times 2.5+0=5\%$

Let x is the resistance per unit length then


equivalent resistance $R=\frac{R_1{R}_2}{R_1+R_2}$

$\frac{\left(x_1{l}_1\right)\left(x_2{l}_2\right)}{x{l}_1+x{l}_2}$

$\Rightarrow \frac{8}{3}=x\frac{l_1}{\frac{l_1}{l_2}+1}$   ...(i)

also $R_0=x{l}_1+x{l}_2$
$12=x(l_1+l_2)$

$12=x{l}_2\left(\frac{l_1}{l_2}+1\right)$   ...(ii)

Now
$\frac{\left(i\right)}{\left(ii\right)}\Rightarrow \frac{\frac{8}{3}}{\frac{12}{1}}=\frac{\frac{x{l}_1}{\left(\frac{l_1}{l_2}+1\right)}}{x{l}_2\left(\frac{l_1}{l_2}+1\right)}$

$=\frac{l_1}{l_2{\left(\frac{l_1}{l_2}+1\right)}^2}$

${\left(\frac{l_1}{l_2}+1\right)}^2\times \frac{8}{36}=\frac{l_1}{l_2}$

$\left(y^2+1+2y\right)\times \frac{8}{36}=y$   (where y = $\frac{l_1}{l_2}$)

8y² + 8 + 16y = 36y

$\Rightarrow$ 8y² – 20y + 8 = 0

$\Rightarrow$ 2y² – 5y + 2 = 0

$\Rightarrow$ 2y² – 4y – y + 2 = 0

$\Rightarrow$ 2y (y – 2) – 1(y – 2) = 0

$\Rightarrow$ (2y – 1) (y – 2) = 0

$\Rightarrow$ $y=\frac{l_1}{l_2}=\frac{1}{2}$ or 2

Time period of cyclotron is

$T=\frac{1}{v}=\frac{2\pi m}{eB};B=\frac{2\pi m}{e}v;R=\frac{mv}{eB}=\frac{p}{eB}$

$\Rightarrow P=eBR=e\times \frac{2\pi mv}{e}R=2\pi mvR$

$K.E.=\frac{p^2}{2m}=\frac{{\left(2\pi mvR\right)}^2}{2m}=2{\pi }^{2}m{v}^2{R}^2$

Magnetic field induction due to vertical loop at the centre O is

$B_1=\frac{{\mu }_{0}I}{2R}$

It acts in horizontal direction.

Magnetic field induction due to horizontal loop at the centre O is

$B_2=\frac{{\mu }_{0}2I}{2R}$

It acts in vertically upward direction.

As B₁ and B₂ are perpendicular to each other, therefore the resultant magnetic field induction at the centre O is

$B_{net}=\sqrt{B_1^2+B_2^2}$

$=\sqrt{{\left(\frac{{\mu }_{0}I}{2R}\right)}^2+{\left(\frac{{\mu }_{0}2I}{2R}\right)}^2}$

$B_{net}=\frac{{\mu }_{0}I}{2R}\sqrt{{\left(1\right)}^2+{\left(2\right)}^2}=\frac{\sqrt{5}{\mu }_{0}I}{2R}$

$S=\frac{V_g}{\left(I-I_g\right)}$

Neglecting I_g

$\therefore$ $S=\frac{V_g}{I}=\frac{25\times {10}^{-3}V}{25A}=0.001\Omega$

A compass needle which is allowed to move in a horizontal plane is taken to a geomagnetic pole. It will stay in any position as the horizontal component of earth’s magnetic field becomes zero at the geomagnetic pole.

According, to Faraday’s law of induction

Induced e.m.f, $\epsilon =-\frac{d\varphi }{dt}$ = -(100t)

Induced current i at t = 2 sec.

= $\left|\frac{\epsilon }{R}\right|=+\frac{100\times 2}{400}$ = +0.5 A

V = -L$\frac{di}{dt}$

$\therefore$ V $\propto$ slope of I-T graph

When L is removed, the phase difference between the voltage and current is

tan$\varphi$₁ = $\frac{X_C}{R}$

$\Rightarrow$ $\tan \frac{\pi }{3}=\frac{X_C}{R}$

$\Rightarrow$ X_C = $\sqrt{3}R$

When C is removed, the phase difference between the voltage and current is

tan$\varphi$₂ = $\frac{X_L}{R}$

$\Rightarrow$ $\tan \frac{\pi }{3}=\frac{X_L}{R}$

$\Rightarrow$ X_L = $\sqrt{3}R$

As X_L = X_C, the series LCR circuit is in resonance.

Net impedence, Z = $\sqrt{R^2+{\left(X_L-X_C\right)}^2}$ = R

Power factor, cos$\varphi$ = $\frac{R}{Z}=\frac{R}{R}=1$

Compare the given equation with

E = E₀cos(kz - $\omega t$)

we get, $\omega$ = 6 $\times$ 10⁸ s^-1

Wave vector, k = $\frac{\omega }{c}$ = $\frac{6\times {10}^8}{3\times {10}^8}$ = 2 m^-1

For normally emerge e = 0

Therefore r₂ = 0 and r₁ = A

Snell’s Law for Incident ray’s

1sin i = $\mu$ sin r₁ = $\mu$sin A

For small angle

i = $\mu$A

According to lens maker's formula

$\frac{1}{f}=\left(\frac{{\mu }_g}{{\mu }_L}-1\right)\left(\frac{1}{R_1}-\frac{1}{R_2}\right)$

As the biconvex lens dipped in a liquid acts as a plane sheet of glass, therefore

f = $\infty$ $\Rightarrow$ $\frac{1}{f}=0$

$\therefore$ $\frac{{\mu }_g}{{\mu }_L}-1=0$

$\Rightarrow$ ${\mu }_g={\mu }_L$

$\therefore$ d = 2f₁ + f₂

Magnifying power, m = $\frac{f_0}{f_e}$ = 9 ..........(1)

where fo and fe are the focal lengths of the objective and eyepiece respectively

Also,f_o + f_e = 20 cm ...(2)

On solving (1) and (2), we get

f_o = 18 cm, f_e = 2 cm

Let, the amount of the two in the mixture will become equal after t years.

The amount of A₁, which remains after t years

N₁ = $\frac{N_{01}}{{\left(2\right)}^{t/20}}$

The amount of A₂, which remains, after t years

N₂ = $\frac{N_{02}}{{\left(2\right)}^{t/10}}$

According to the question

N₁ = N₂

$\Rightarrow$ $\frac{40}{{\left(2\right)}^{t/20}}=\frac{160}{{\left(2\right)}^{t/10}}$

$\Rightarrow$ ${\left(2\right)}^{t/20}$ = ${\left(2\right)}^{\left(\frac{t}{10}-2\right)}$

$\Rightarrow$ $\frac{t}{20}=\left(\frac{t}{10}-2\right)$

$\Rightarrow$ t = 40 s

We know,

$\frac{1}{\lambda }=R{Z}^2\left(\frac{1}{n_1^2}-\frac{1}{n_2^2}\right)$

where R = Rydberg constant, Z = atomic number

Here, n₁ = 1, n₂ = 5

$\therefore$ $\frac{1}{\lambda }=R\left(\frac{1}{1^2}-\frac{1}{5^2}\right)$ = $\frac{24R}{25}$

According to conservation of linear momentum, we get

Momentum of photon = Momentum of atom

$\Rightarrow$ $\frac{h}{\lambda }$ = mv

$\Rightarrow$ v = $\frac{h}{m\lambda }$ = $\frac{h}{m}\left(\frac{24R}{25}\right)$ = $\frac{24hR}{25m}$

Nuclear radius, R = R₀A^1/3

where R₀ is a constant and A is the mass number

$\therefore$ $\frac{R_{Al}}{R_{Cu}}=\frac{{\left(27\right)}^{1/3}}{{\left(64\right)}^{1/3}}$ = $\frac{3}{4}$

$\Rightarrow$ R_Cu = $\frac{4}{3}\times R_{Al}$ = $\frac{4}{3}\times 3.6$ = 4.8 fermi

In first case,

n₁ = 3 and n₂ = 4

$\frac{1}{{\lambda }_1}=R\left(\frac{1}{3^2}-\frac{1}{4^2}\right)$ = $\frac{7R}{144}$

In second case,

n₁ = 2 and n₂ = 3

$\frac{1}{{\lambda }_2}=R\left(\frac{1}{2^2}-\frac{1}{3^2}\right)$ = $\frac{5R}{36}$

$\therefore$ $\frac{{\lambda }_1}{{\lambda }_2}$ = $\frac{5}{36}\times \frac{144}{7}$ = $\frac{20}{7}$

Wavelength $\lambda$ = $\frac{h}{p}$

$\Rightarrow$ $\frac{h}{p}$ = $\frac{h}{mv}$

For circular motion = F_c = qvB

$\Rightarrow$ $\frac{m{v}^2}{r}$ = qvB

$\Rightarrow$ mv = rBq

$\Rightarrow$ mv = (0.83 × 10^–2)(0.25)(2 × 1.6 × 10^–19)

de Broglie wavelength,

$\lambda$ = $\frac{h}{mv}$

= $\frac{6.6\times {10}^{-34}}{0.83\times {10}^{-2}\times 0.25\times 2\times 1.6\times {10}^{-19}}$

= 0.01 $\stackrel{\circ }{A}$

Energy released from first exited state to the ground state :

E = (– 3.4) – (– 13.6) = 10.2 eV

Work function $\varphi$ = E – 3.75 eV = (10.2 – 3.75) eV

= 6.63 eV

$\Rightarrow$ h${\nu }_0$ = 6.63

$\Rightarrow$ ${\nu }_0$ = $\frac{6.63\times 1.6\times {10}^{-19}}{6.63\times {10}^{-34}}$ = 1.6 $\times$ 10¹⁵ Hz

Efficient power, P = $\frac{N}{t}\times \frac{hc}{\lambda }$

$\Rightarrow$ $\frac{N}{t}$ = $\frac{\frac{25}{100}\times 200}{\frac{\left(6.6\times {10}^{-34}\right)\left(3\times {10}^8\right)}{0.6\times {10}^{-6}}}$ = 1.5 $\times$ 10²⁰

Here D₁ is in forward bias and D₂ is in reverse bias so, D₁ will conduct and D₂ will not conduct.
So, the current supplied by the battery is

I = $\frac{5}{10}$ = 0.5 A

Here, R_C = 2 k$\Omega$ = 2 × 10³ $\Omega$

V₀ = 2 V, R_B = 1 k$\Omega$ = 1 × 10³ $\Omega$, $\beta$ = 100

The output voltage, across the load R_C

V₀ = I_CR_C = 2

The collector current (I_C)

I_C = $\frac{2}{2\times {10}^3}$ = 10^-3 A = 1 mA

Current gain($\beta$) = $\frac{I_C}{I_B}$ = 100

$\Rightarrow$ I_B = $\frac{I_C}{100}$ = $\frac{{10}^{-3}}{100}$ = 10^-5 A

Input voltage, V_i

= I_BR_B = (10^–5 A) (1 × 10³ $\Omega$) = 10^–2 V = 10 mV

In the given graph,

Region (I) – Cutoff region

Region (II) – Active region

Region (III) – Saturation region

Using transistor as a switch it is used in cutoff region or saturation region.

Using transistor as a amplifier it is used in active region.

The truth table is
The logic circuit is OR gate.

Electronic configuration of ⁶C

⁶C = 1s², 2s² 2p²

The electronic configuration of ¹⁴Si

¹⁴Si = 1s², 2s² 2p⁶, 3s² 3p²

As they are away from Nucleus, so effect of nucleus is low for Si even for Sn and Pb are almost mettalic.

Rb(37) = 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁶ 5s¹

For 5s, n= 5, l = 0, m = 0, s = +1/2 or -1/2

l = 3 and n = 4 represent 4f, So total number of electrons = 2(2$l$+1) = 14

This is an example of disproportionation reaction and oxidation state of chlorine changes from 0 to –1 and +5.

KClO₃ + C₂H₂O₄ + H₂SO₄ $\to$ K₂SO₄ + CO₂ + KCl + H₂O

Maximum change in oxidation number of chlorine, i.e., from +5 to –1.

According to Graham's law of diffusion,

$\frac{r_1}{r_2}=\sqrt{\frac{d_2}{d_1}}=\sqrt{\frac{M_2}{M_1}}$

$r_A=\frac{V_A}{T_A},r_B=\frac{V_B}{T_B}$

$\frac{V_A/T_A}{V_B/T_B}=\sqrt{\frac{M_B}{M_A}}$

$V_A=V_B,T_A=150\sec$, T_B = 200 sec, M_B = 36, M_A = ?

$\frac{T_B}{T_A}=\sqrt{\frac{M_B}{M_A}}\Rightarrow \frac{200}{150}=\sqrt{\frac{36}{M_A}}$

$\Rightarrow \frac{4}{3}=\sqrt{\frac{36}{M_A}}or\frac{4\times 4}{3\times 3}=\frac{36}{M_A}$

$\Rightarrow$ M_A = $\frac{36}{4\times 4}\times 3\times 3=20.25$

All of the given salts have same anion i.e., Cl which on hydrolysis gives HCl which is a strong acid.

Now, among the salts which have cation that gives a strongest base on hydrolysis of salt have the highest pH value. As Ba form Ba(OH)₂ which is a stronger base thus, it results in the highest pH value.

For this example,

CH₃COOH ⇌ CH₃COO^– + H⁺ ;

CH₃COONa ⇌ CH₃COO^– + Na⁺

when few drops of HCl are added to this buffer, the H⁺ of HCl immediatly combine with CH₃COO^– ions to form undissociated acetic acid molecules. Thus there will be no H⁺ ions to combine with CH₃COO^– ions to form undissociated acetic acid molecules. Thus there will be no appreciable change in its pH value. Like wise if few drops of NaOH are added, the OH^– ions will combine with H⁺ ions to form unionised water molecule. Thus pH of solution will remain constant.

	Ba(OH)2	⇌	Ba2+	+	2OH
At equilibrium			x		2x

pH = – log[H⁺]

12 = – log [H⁺]

$\Rightarrow$ [H⁺] = 10^–12

As, [H⁺][OH^– ] = 10^–14

10^–12 [OH^– ] = 10^–14

$\Rightarrow$ [OH^– ] = 10^–2

As [OH^– ] = 2x = 10^–2 then x = 5.0 × 10^–3

Now, K_sp = [Ba²⁺][OH^– ]²

K_sp = (5 × 10^–3) (10^–2)² = 5.0 × 10^–7

p = p_Ax_A + p_Bx_B

= p_Ax_A + p_B(1 – x_A)

(As for binary solution x_A + x_B = 1)

= p_Ax_A + p_B – p_Bx_A

= p_B + x_A(p_A – p_B)

H₂O(l) $\frac{\text{buildrel}100^{\circ }C}{⟶}$ H₂O(g)

∆H^o = 40.66kJ mol^–1

∆H^o = ∆u^o + $\Delta$n_g RT

$\Delta$n_g = 1, R = 8.314 × 10^–3 kJ mol^–1 k^–1

T = 100 + 273 = 373 K

$\Rightarrow$ 40.66 = ∆u^o + (1) (8.314 × 10^–3) × 373

$\Rightarrow$ ∆u^o = 37.56 kJ mol^–1

H₂O($l$) → H₂O(s)

∆H = 1.435 Kcal/mol

T = 0 + 273K = 273K

$\Delta S=\frac{\Delta H}{T}$

$\Rightarrow$ $\Delta S=\frac{1.435}{273}$ = 5.26 $\times$ 10^-3 kcal/mol K

$\Rightarrow$ $\Delta S$ = 5.260 cal/mol K