Tag: gases and the kinetic theory

In this case, Initial pressure of oxygen $=$ 1 atm and initial volume of oxygen $=$ 1 lit
Initial pressure of nitrogen $=$ 0.5 atm and inititial volume of nitrogen $=$ 2 lit
Since temperature is constant, pressure (P) and volume (V) of the combined gas will be
$PV = P _{oxygen}V _{oxygen} + P _{nitrogen}V _{nitrogen}$
Given, volume of combined gas $=$ 1 lit.
${P}{(1)}={(1)}{(1)}+{(0.5)}{(2)}$
${P} ={2}$ atm
Putting all these values we get pressure of the combined gas $=$ 2 atm

Since on raising the tube, the volume of vacuum will decrease and hence pressure due to alcohol will decrease. Since the pressure at the base must be equal to atmospheric pressure, to compensate for the decrease in pressure due to alcohol expansion, mercury in tube will rise but will never reach 76mm since pressure due to vaporized alcohol will never be 0.

Assuming the process to be isothermal, we can sa for an ideal gas,

If initial volume = v, final volume = $\dfrac{95v}{100}$
If P1 = initial pressure, P2 (final pressure) =$\dfrac{100P _1}{95}$
So increase in pressure = $\dfrac{100P _1}{95 - P _1} = \dfrac{5P _1}{95}$
So, percentage increase in pressure = $5\times \dfrac{100}{95} = 5.26%$

${ P } _{ 1 }{ V } _{ 1 }={ P } _{ 2 }{ V } _{ 2 }\ First\quad increased\quad by\quad 50\quad i.e\quad { P } _{ 2 }=1.5{ P } _{ 1 }\ { V } _{ 2 }=\dfrac { 1 }{ 1.5 } { V } _{ 1 }\ Now\quad decreased\quad by\quad 50\quad i.e\quad { P } _{ 3 }=\dfrac { { P } _{ 2 } }{ 2 } =\dfrac { 1.5 }{ 2 } { P } _{ 1 }\ { P } _{ 2 }{ V } _{ 2 }={ P } _{ 3 }V _{ 3 }\ V _{ 3 }=2{ V } _{ 2 }\ V _{ 3 }=\dfrac { 2 }{ 1.5 } { V } _{ 1 }\ V _{ 3 }=\dfrac { 20 }{ 15 } { V } _{ 1 }=\dfrac { 4 }{ 3 } { V } _{ 1 }$

We know,
P1V1 = P2V2
At the top of the lake P1 = $P _o$ (say)
If h = height of the lake then,
($P _o$ + $\rho g h ) (4/3) \pi r^3$= $P _o$ $(4/3) 8 \pi r^3$
Solving we get, h = 7$P _0$/ $\rho$ g
Now, $P _o$ = 1 X 9.8 X 11
Considering density of water = 1,
we get, h = 11 x 7 = 77 m

In this case, (4/3) $\pi r^3(p _0 +p _1)$ = (4/3) $\pi r^3 (5^3/4^3)P _0$
where, $P _0$ = atm pressure, $P _1$ = water pressure
Rearranging, we get, $P _1 = [(5^3/4^3) - 1] P _0$
Or, $\rho g H = (61/64) P _0$
Or, H = $(61/64) P _0/\rho g$
$P _0 $= 10g and $\rho$ of water = 1
Putting these values we get, H = 9.53m

As process takes place at constant temperature, therefore, when air bubble rises upwards pressure decreases therefore volume of air bubble will definitely increase.
Assertion is true and reason is correct explaination of it.
option(A) is correct

At a depth of h water pressure will be
${P} _{h} = {P} _{a} + \rho gh$                eq(1)
where 
$\rho  =1000 kg/{m}^{3}$
$g  = 9.81 m/{s}^{2} $
$h = \text{depth  of  water  bubble}$
${P} _{a} = {10}^{5}Pa$

initially bubble is below water at h, so
${P} _{1} = {P} _{h}$
finally it rises to surface of lake so
${P} _{2} = {P} _{a}$

we assume that temperature is constant when it is rising from bottom to surface
by ideal gas equation
${P} _{1}{V} _{1} = {P} _{2}{V} _{2}$

volume can be written as $V = \dfrac{4}{3}\pi {r}^{3}$
where r is radius.
${P} _{h} \times \dfrac{4}{3}\pi {r} _{h}^{3} = {P} _{a} \times \dfrac{4}{3}\pi {r} _{s}^{3}$
${r} _{h} = {(\dfrac{{P} _{a}}{{P} _{h}})}^{1/3} r$
${r} _{h} = \dfrac{r}{{(\frac{{P} _{h}}{{P} _{a}})}^{1/3}}$
by eq(1)
${r} _{h} = \dfrac{r}{{(\dfrac{{P} _{a} +  \rho gh}{{P} _{a}})}^{1/3}}$
${r} _{h} = \dfrac{r}{{(1 + 0.0981h)}^{1/3}}$              eq(2)

(a) at depth 30m
h = 30m
put h=30 in eq(2)
${r} _{h} = \dfrac{r}{1.5}$
false

(b) at depth 70m
h = 70m
put h=70 in eq(2)
${r} _{h} = \dfrac{r}{2}$
true

(c) at depth 140m
h = 140m
put h=140 in eq(2)
${r} _{h} = \dfrac{r}{2.4}$
false

(d) at depth 260m
h = 260m
put h=260 in eq(2)
${r} _{h} = \frac{r}{3}$
true

Answer is D.