NEET-XII-Physics
24: Kinetic Theory of Gases
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- Qstn #29A vessel contains 1.60 g of oxygen and 2.80 g of nitrogen. The temperature is maintained at 300 K and the volume of the vessel is 0.166 m3. Find the pressure of the mixture.Ans : `` \begin{array}{l}\text{Here,}\\ \text{V = 0}{\text{.166 m}}^{3}\\ T=300K\\ {\text{Mass of O}}_{2}=1.60g\\ {M}_{O}=32g\\ {n}_{O}=\frac{1.60}{32}=0.05\\ {\text{Mass of N}}_{2}=2.80g\\ {M}_{N}=28g\\ {n}_{N}=\frac{2.80}{28}=0.1\\ {\text{Partial pressure of O}}_{2}\text{is given by}\\ {P}_{O}=\frac{{n}_{O}RT}{V}=\frac{0.05\times 8.3\times 300}{0.166}=750\\ {\text{Partial pressure of N}}_{2}\text{is given by}\\ {P}_{N}=\frac{{n}_{N}RT}{V}=\frac{0.1\times 8.3\times 300}{0.166}=1500\\ \text{Total pressure is sum of the partial pressures.}\\ \Rightarrow P={P}_{N}+{P}_{O}=750+1500=2250Pa\end{array}``
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- Qstn #30A vertical cylinder of height 100 cm contains air at a constant temperature. The top is closed by a frictionless light piston. The atmospheric pressure is equal to 75 cm of mercury. Mercury is slowly poured over the piston. Find the maximum height of the mercury column that can be put on the piston.Ans : `` \begin{array}{l}\text{Here,}\\ h=1m\\ \\ {P}_{1}=0.75\text{mHg = 0}\text{.75}\rho \text{g Pa}\\ \rho {\text{=13500 kg/m}}^{3}\\ \text{Let h be the height of the mercury above the piston.}\\ {P}_{2}={P}_{1}+h\rho g\\ \text{Let the CSA be A.}\\ {V}_{1}=Ah=A\\ {V}_{2}=(1-h)A\\ \text{Applying Boyle's law, we get}\\ {P}_{1}{V}_{1}={P}_{2}{V}_{2}\\ \Rightarrow 0.75\rho gA={P}_{2}(1-h)A\\ \Rightarrow 0.75\rho g=(0.75\rho g+h\rho g)(1-h)\\ \Rightarrow 0.75=(0.75+h)(1-h)\\ \Rightarrow h=0.25\text{m}\\ \text{h = 25 cm}\end{array}``
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- Qstn #31Figure shows two vessels A and B with rigid walls containing ideal gases. The pressure, temperature and the volume are pA, TA, V in the vessel A and pB, TB, V in the vessel B. The vessels are now connected through a small tube. Show that the pressure p and the temperature T satisfy
pT=12pATA+pBTBwhen equilibrium is achieved.
Figure
Ans : `` \begin{array}{l}\\ {\text{Let the partial pressure of the gas in chamber A and B be P}}_{A}^{\text{'}}{\text{and P}}_{B}^{\text{'}}\text{, respectively.}\\ \\ \text{Applying equation of state for gas A, we get}\\ \frac{{P}_{A}V}{{T}_{A}}=\frac{{P}_{A}^{\text{'}}2V}{T}\\ \Rightarrow {P}_{A}^{\text{'}}=\frac{{P}_{A}T}{2{T}_{A}}\\ \text{Similarly, for gas B:}\\ {P}_{B}^{\text{'}}=\frac{{P}_{B}T}{2{T}_{B}}\\ \text{Total pressure is the sum of the partial pressures. It is given by}\\ P={P}_{A}^{\text{'}}+{P}_{B}^{\text{'}}\\ =\frac{{P}_{A}T}{2{T}_{A}}+\frac{{P}_{B}T}{2{T}_{B}}\\ \Rightarrow P=\frac{T}{2}(\frac{{P}_{A}}{{T}_{A}}+\frac{{P}_{B}}{{T}_{B}})\\ \Rightarrow \frac{P}{T}=\frac{1}{2}(\frac{{P}_{A}}{{T}_{A}}+\frac{{P}_{B}}{{T}_{B}})\end{array}``
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- Qstn #32A container of volume 50 cc contains air (mean molecular weight = 28.8 g) and is open to atmosphere where the pressure is 100 kPa. The container is kept in a bath containing melting ice (0°C).
a) Find the mass of the air in the container when thermal equilibrium is reached.
(b) The container is now placed in another bath containing boiling water (100°C). Find the mass of air in the container.
(c) The container is now closed and placed in the melting-ice bath. Find the pressure of the air when thermal equilibrium is reached.Ans : `` \begin{array}{l}\left(\text{a}\right)\text{}\text{Here,}\\ {V}_{1}=5\times {10}^{-5}{m}^{3}\\ {P}_{1}={10}^{5}\text{Pa}\\ {T}_{1}=273\text{K}\\ \text{M = 28}\text{.8 g}\\ {P}_{1}{V}_{1}=nR{T}_{1}\\ \Rightarrow n=\frac{{P}_{1}{V}_{1}}{R{T}_{1}}\\ \Rightarrow \frac{m}{M}=\frac{{10}^{5}\times 5\times {10}^{-5}}{8.3\times 273}\\ \Rightarrow m=\frac{{10}^{5}\times 5\times {10}^{-5}\times 28.8}{8.3\times 273}\\ \Rightarrow m=0.0635\text{g}\end{array}``
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`` \begin{array}{l}\left(\text{b}\right)\,\mathrm{\,Here\,},\\ {V}_{1}=5\times {10}^{-5}{m}^{3}\\ {P}_{1}={10}^{5}\text{Pa}\\ {P}_{2}={10}^{5}\text{Pa}\\ {T}_{1}=273\text{K}\\ {T}_{2}=373\text{K}\\ \text{M = 28}\text{.8 g}\\ \frac{{P}_{1}{V}_{1}}{{T}_{1}}=\frac{{P}_{2}{V}_{2}}{{T}_{2}}\\ \Rightarrow \frac{\text{}5\times {10}^{-5}}{273}=\frac{{V}_{2}}{373}\\ \Rightarrow {V}_{2}=\frac{\text{}5\times {10}^{-5}\times 373}{273}\\ \Rightarrow {V}_{2}=6.831\times {10}^{-5}\\ \text{Volume of expelled air =}6.831\times {10}^{-5}-5\times {10}^{-5}\\ =1.831\times {10}^{-5}\\ \text{Applying equation of state, we get}\\ PV=nRT\\ \Rightarrow \frac{m}{M}=\frac{PV}{RT}=\frac{{10}^{5}\times 1.831\times {10}^{-5}}{8.3\times 373}\\ \Rightarrow m=\frac{28.8\times {10}^{5}\times 1.831\times {10}^{-5}}{8.3\times 373}=0.017\\ \text{Thus, mass of expelled air = 0}\text{.017 g}\\ \text{Amount of air in the container = 0}\text{.0635}-\text{0}\text{.017}\\ =0.0465\text{g}\end{array}``
`` \begin{array}{l}\left(\text{c}\right)\text{Here,}\\ T=273K\\ P={10}^{5}\text{Pa}\\ \text{V=5}\times {\text{10}}^{-5}{\text{m}}^{3}\\ \text{Applying equation of state, we get}\\ PV=nRT\\ \Rightarrow P=\frac{nRT}{V}=\frac{0.0465\times 8.3\times 273}{28.8\times 5\times {10}^{-5}}\\ \Rightarrow P=0.731\times {10}^{5}\approx 73\text{kPa}\end{array}``
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- Qstn #33A uniform tube closed at one end, contains a pellet of mercury 10 cm long. When the tube is kept vertically with the closed-end upward, the length of the air column trapped is 20 cm. Find the length of the air column trapped when the tube is inverted so that the closed-end goes down. Atmospheric pressure = 75 cm of mercury.Ans : `` \begin{array}{l}\text{Let the CSA of the tube be A.}\\ {\text{Initial volume of air, V}}_{1}\text{= 20A cm = 0}\text{.2A}\\ \text{Length of mercury, h = 0}\text{.1 m}\\ \text{}\\ \,\mathrm{\,Let\; the\,}\text{pressure of the trapped}\,\mathrm{\,air\,}\text{w}\,\mathrm{\,hen\; the\; tube\; is\; inverted\; and\; vertical\,}{\text{be P}}_{1}\text{.}\\ \text{Now, pressure of the mercury and trapped air}\,\mathrm{\,balances\,}\,\mathrm{\,the\,}\,\mathrm{\,atmospheric\; pressure\,}.\,\mathrm{\,Thus\,},\text{}\\ \\ {P}_{1}+0.1\rho g=0.75\rho g\\ \Rightarrow {P}_{1}=0.65\rho g\\ \text{When the tube is inverted with the closed end down, the pressure acting upon the trapped air is given by}\\ \\ \text{Atmospheric pressure + Mercury column pressure}\\ \,\mathrm{\,Now\,},\\ \text{Pressure of trapped air = Atmospheric pressure + Mercury column pressure}\left[\text{In equilibrium}\right]\\ {P}_{2}=0.75\rho g+0.1\rho g=0.85\rho g\\ \text{Applying the Boyle's law when the t}\,\mathrm{\,emperature\; remains\,}constant,\,\mathrm{\,we\,}\,\mathrm{\,get\,}\\ {P}_{1}{V}_{1}={P}_{2}{V}_{2}\\ \text{Let the new height of the trapped air be x.}\\ \Rightarrow 0.65\rho g\text{0}\text{.2A =}0.85\rho gxA\\ \Rightarrow x=0.15\text{m = 15 cm}\end{array}``
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- Qstn #34A glass tube, sealed at both ends, is 100 cm long. It lies horizontally with the middle 10 cm containing mercury. The two ends of the tube contain air at 27°C and at a pressure 76 cm of mercury. The air column on one side is maintained at 0°C and the other side is maintained at 127°C. Calculate the length of the air column on the cooler side. Neglect the changes in the volume of mercury and of the glass.Ans : `` \begin{array}{l}\text{Let CSA of the tube be A.}\\ \text{On the colder side:}\\ {P}_{1}=0.76\text{m Hg}\\ {T}_{1}=300K\\ {V}_{1}=V\text{}\\ {T}_{2}=273K\\ {V}_{2}=Ax\\ \frac{{P}_{1}V}{{T}_{1}}=\frac{{P}_{2}Ax}{{T}_{2}}\\ \Rightarrow {P}_{2}=\frac{{P}_{1}V{T}_{2}}{{T}_{1}Ax}\\ \text{On the hotter side:}\\ {P}_{1}=0.76\text{m Hg}\\ {T}_{1}=300K\\ {{V}_{1}}^{\text{'}}=V\text{}\\ {{T}_{2}}^{\text{'}}=400K\\ {{V}_{2}}^{\text{'}}=Ay\\ \frac{{{P}_{1}}^{\text{'}}V}{{T}_{1}}=\frac{{{P}_{2}}^{\text{'}}Ay}{{{T}_{2}}^{\text{'}}}\\ \Rightarrow {{P}_{2}}^{\text{'}}=\frac{{P}_{1}V{{T}_{2}}^{\text{'}}}{{T}_{1}Ay}\\ \text{In equilibrium, the pressures on both side will balance each other.}\\ \Rightarrow {{P}_{2}}^{\text{'}}={P}_{2}\\ \Rightarrow \frac{{P}_{1}V{{T}_{2}}^{\text{'}}}{{T}_{1}Ay}=\frac{{P}_{1}V{T}_{2}}{{T}_{1}Ax}\\ \Rightarrow \frac{{{T}_{2}}^{\text{'}}}{y}=\frac{{T}_{2}}{x}\\ \text{From the length of the tube, we get}\\ x+y+0.1=1\\ \Rightarrow y=0.9-x\\ \frac{400}{(0.9-x)}=\frac{273}{x}\\ \Rightarrow x=0.365\text{m}\\ \Rightarrow \text{}x=36.5\text{cm}\end{array}``
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- Qstn #35An ideal gas is trapped between a mercury column and the closed-end of a narrow vertical tube of uniform base containing the column. The upper end of the tube is open to the atmosphere. The atmospheric pressure equals 76 cm of mercury. The lengths of the mercury column and the trapped air column are 20 cm and 43 cm respectively. What will be the length of the air column when the tube is tilted slowly in a vertical plane through an angle of 60°? Assume the temperature to remain constant.Ans :
`` \begin{array}{l}\text{Here,}\\ \text{Initial pressure = Atmospheric pressure + Pressure due to mercury}\\ \\ \text{⇒}{P}_{1}={P}_{o}+{P}_{Hg}\\ \text{Let the CSA of the tube be A.}\\ {P}_{1}=0.76+0.2\text{=0}\text{.96 m Hg}\\ {T}_{1}={T}_{2}=T\\ {V}_{1}=0.43A\\ \text{}{\text{If the tube is slanted, then the atmospheric pressure P}}_{o}\text{}{\text{remains the same. Only the P}}_{Hg}\text{changes}\text{.}\\ \\ {P}_{2}={P}_{o}+{P}_{Hg}\,\mathrm{\,cos\,}{60}^{0}=0.76+0.2\times 0.5=0.86\text{}\\ {P}_{1}{V}_{1}={P}_{2}{V}_{2}\\ \Rightarrow {V}_{2}=\frac{{P}_{1}{V}_{1}}{{P}_{2}}=\frac{\text{0}\text{.96}\times 0.43A}{0.86\text{}}\\ \text{Let the length of the air column be}l.\\ \Rightarrow Al=\frac{{P}_{1}{V}_{1}}{{P}_{2}}=\frac{\text{0}\text{.96}\times 0.43A}{0.86\text{}}\\ \Rightarrow l=0.48\text{m}\\ \Rightarrow l=48\text{cm}\end{array}``
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- Qstn #36Figure shows a cylindrical tube of length 30 cm which is partitioned by a tight-fitting separator. The separator is very weakly conducting and can freely slide along the tube. Ideal gases are filled in the two parts of the vessel. In the beginning, the temperatures in the parts A and B are 400 K and 100 K respectively. The separator slides to a momentary equilibrium position shown in the figure. Find the final equilibrium position of the separator, reached after a long time.
Figure
Ans : Let the initial pressure of the chambers A and B be PA1 and PB1, respectively.
Let the final pressure of chambers A and B be PA2 and PB2, respectively.
`` \begin{array}{l}\text{}\\ \text{Let the CSA be A.}\\ {V}_{A1}=0.20A\\ {T}_{A1}=400\text{K}\\ {V}_{B1}=0.1A\\ {T}_{B1}=100\text{K}\\ \text{At first equilibrium, both side pressures will be the same.}\\ \Rightarrow {P}_{A1}={P}_{B1}\\ \text{Let the final temperature at equilibrium be T. Then,}\\ \frac{{P}_{A1}{V}_{A1}}{{T}_{A1}}=\frac{{P}_{A2}{V}_{A2}}{T}\\ \Rightarrow \frac{{P}_{A1}0.2A}{400}=\frac{{P}_{A2}{V}_{A2}}{T}\\ \Rightarrow {P}_{A2}=\frac{{P}_{A1}0.2AT}{400{V}_{A2}}...\left(1\right)\\ \text{For second chamber:}\\ \frac{{P}_{B1}{V}_{B1}}{{T}_{B1}}=\frac{{P}_{B2}{V}_{B2}}{T}\\ \Rightarrow \frac{{P}_{B1}0.1A}{100}=\frac{{P}_{B2}{V}_{B2}}{T}\\ \Rightarrow {P}_{B2}=\frac{{P}_{B1}0.1AT}{100{V}_{B2}}...\left(2\right)\\ \text{At second equilibrium, pressures on both sides will be the same.}\\ \Rightarrow {P}_{A2}={P}_{B2}\\ \Rightarrow \frac{{P}_{A1}0.2AT}{400{V}_{A2}}=\frac{{P}_{B1}0.1AT}{100{V}_{B2}}\\ \Rightarrow \frac{{P}_{A1}}{2{V}_{A2}}=\frac{{P}_{B1}}{{V}_{B2}}\\ \text{⇒}{V}_{B2}=2{V}_{A2}...\left(3\right)\\ \text{Now,}\\ {V}_{B2}+{V}_{A2}=0.3A\\ \Rightarrow 3{V}_{A2}=0.3A\\ \Rightarrow {V}_{A2}=0.1A\\ \text{Let}{V}_{A2}\,\mathrm{\,be\,}lA.\\ \Rightarrow l=0.1\text{m}\\ \text{⇒}l=10\text{cm}\end{array}``
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- Qstn #37A vessel of volume V0 contains an ideal gas at pressure p0 and temperature T. Gas is continuously pumped out of this vessel at a constant volume-rate dV/dt = r keeping the temperature constant. The pressure of the gas being taken out equals the pressure inside the vessel. Find
a) the pressure of the gas as a function of time,
(b) the time taken before half the original gas is pumped out.Ans : `` \begin{array}{l}\text{Let}\mathit{\text{P}}\text{be the pressure and}\mathit{\text{n}}\text{be the number of moles of gas inside the}\,\mathrm{\,vessel\; at\; any\; given\; time\,}t.\text{}\\ \text{}\\ \text{Suppose a small amount of gas of}\mathit{\text{dn}}\text{moles is pumped out and the decrease in pressure is}\mathit{\text{dP.}}\\ \\ \text{Applying equation of state to the gas inside the vessel, we get}\\ (P-dP){V}_{o}=(n-dn)RT\\ \Rightarrow P{V}_{o}-dP{V}_{o}=nRT-dnRT\\ \text{But}P{V}_{o}=nRT\\ \Rightarrow {V}_{o}dP=dnRT...\left(1\right)\\ \text{The pressure of the gas taken out is equal to the inner pressure}\text{.}\\ \text{Applying equation of state, we get}\\ (P-dP)dV\text{=}dnRT\\ \Rightarrow PdV=dnRT...\left(2\right)\\ \text{From eq.}\left(1\right)\text{and eq.}\left(2\right)\text{, we get}\\ {V}_{o}dP=PdV\\ \Rightarrow \frac{dP}{P}=\frac{dV}{{V}_{o}}\\ \frac{dV}{dt}=r\\ \Rightarrow dV=rdt\\ \\ \Rightarrow dV=-rdt...\left(3\right)\left[\,\mathrm{\,Since\,}\,\mathrm{\,pressures\,}\,\mathrm{\,decreases\,},\,\mathrm{\,rate\,}\,\mathrm{\,is\,}\,\mathrm{\,negative\,}\right]\\ \text{Now,}\\ \frac{dP}{P}=\frac{-rdt}{{V}_{o}}\left[\,\mathrm{\,From\,}\,\mathrm{\,eq\,}.\left(3\right)\right]\\ \left(a\right)\\ {\text{Integrating the equation P = P}}_{0}\text{to P = P and time t = 0 to t = t, we get}\\ \underset{{P}_{o}}{\overset{P}{\int }}=\underset{0}{\overset{t}{\int }}\\ \Rightarrow \,\mathrm{\,ln\,}P-\,\mathrm{\,ln\,}{P}_{o}=-\frac{rt}{{V}_{o}}\\ \Rightarrow \,\mathrm{\,ln\,}\left(\frac{P}{{P}_{o}}\right)=-\frac{rt}{{V}_{o}}\\ \Rightarrow P={P}_{o}{e}^{\frac{-rt}{{V}_{o}}}\end{array}``
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`` \begin{array}{l}\left(b\right)\\ \text{P =}\frac{{P}_{o}}{2}\\ \frac{{P}_{o}}{2}={P}_{o}{e}^{\frac{-rt}{{V}_{o}}}\\ \Rightarrow {e}^{\frac{rt}{{V}_{o}}}=2\\ \Rightarrow \frac{rt}{{V}_{o}}=\,\mathrm{\,ln\,}2\\ \Rightarrow t=\frac{{V}_{o}\,\mathrm{\,ln\,}2}{r}\end{array}``
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- Qstn #38One mole of an ideal gas undergoes a process
p=p01+(V/V0)2where p0 and V0 are constants. Find the temperature of the gas when V = V0.Ans : `` \begin{array}{l}\text{Given:}\\ p=\frac{{p}_{o}}{1+{\left(\frac{V}{{V}_{o}}\right)}^{2}}\\ \text{Multiplying both sides by V, we get}\\ pV=\frac{{p}_{o}V}{1+{\left(\frac{V}{{V}_{o}}\right)}^{2}}\\ \\ pV=RT\left[\,\mathrm{\,From\,}\,\mathrm{\,eq\,}.\left(1\right)\right]\\ \text{Now,}\\ RT=\frac{{p}_{o}V}{1+{\left(\frac{V}{{V}_{o}}\right)}^{2}}\\ \\ T=\frac{1}{R}\left(\frac{{p}_{o}{V}_{o}}{1+{\left(\frac{{V}_{o}}{{V}_{o}}\right)}^{2}}\right)\left[{\text{V=V}}_{o}\right]\\ \Rightarrow T=\frac{{p}_{o}{V}_{o}}{2R}\end{array}``
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- Qstn #39Show that the internal energy of the air (treated as an ideal gas) contained in a room remains constant as the temperature changes between day and night. Assume that the atmospheric pressure around remains constant and the air in the room maintains this pressure by communicating with the surrounding through the windows, doors, etc.Ans : `` \begin{array}{l}\text{We know internal energy at a particular temperature}\\ U=n{C}_{v}T\\ \text{Air in home is are chiefly diatomic molecules, so}\\ {C}_{v}\text{=}\frac{5}{2}R\text{}\\ \therefore U=\frac{5n}{2}RT\\ \text{Now by eqn}\text{. of state}\\ nRT=PV\\ U=\frac{5}{2}\left(nRT\right)\\ =>U=\frac{5}{2}PV\\ \text{Now pressure P is constant also V of the room = constant}\\ \text{Thus,}\\ U=\text{constant}\end{array}``
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- Qstn #40Figure shows a cylindrical tube of radius 5 cm and length 20 cm. It is closed by a tight-fitting cork. The friction coefficient between the cork and the tube is 0.20. The tube contains an ideal gas at a pressure of 1 atm and a temperature of 300 K. The tube is slowly heated and it is found that the cork pops out when the temperature reaches 600 K. Let dN denote the magnitude of the normal contact force exerted by a small length dl of the cork along the periphery (see the figure). Assuming that the temperature of the gas is uniform at any instant, calculate
dNdl.
Figure
Ans : `` \begin{array}{l}\text{Here,}\\ {P}_{1}={10}^{5}\text{Pa}\\ A=\pi {\left(0.05\right)}^{2}\\ L=0.2\text{m}\\ V=AL=0.0016{\text{m}}^{3}\\ {T}_{1}=300\text{K}\\ {T}_{2}=600\text{K}\\ \mu \text{=0}\text{.20}\\ \text{Applying 5 variable equation of state, we get}\\ \frac{{P}_{1}V}{{T}_{1}}=\frac{{P}_{2}V}{{T}_{2}}\\ \Rightarrow \frac{{P}_{1}}{{T}_{1}}=\frac{{P}_{2}}{{T}_{2}}\\ \Rightarrow {P}_{2}=\frac{{T}_{2}}{{T}_{1}}\times {P}_{1}=\frac{600}{300}\times {10}^{5}\\ \Rightarrow {P}_{2}=2\times {10}^{5}\\ \text{Net pressure, P =}{P}_{2}-{P}_{1}\text{=}2\times {10}^{5}-{10}^{5}={10}^{5}\\ \text{Total force acting on the stopper =}PA{\text{=10}}^{5}\times \pi \times {\left(0.05\right)}^{2}\\ \text{Applying law of friction, we get}\\ F=\mu N=0.2N\\ \Rightarrow N=\frac{F}{\mu }=\frac{{\text{10}}^{5}\times \pi \times {\left(0.05\right)}^{2}}{0.2}\\ \frac{dN}{dl}=\frac{N}{2\pi r}=\frac{{\text{10}}^{5}\times \pi \times {\left(0.05\right)}^{2}}{0.2\times 2\pi \times \left(0.05\right)}=0.125\times {10}^{5}\\ \Rightarrow \frac{dN}{dl}=1.25\times {10}^{4}\text{ N/m}\end{array}``
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- Qstn #41Figure shows a cylindrical tube of cross-sectional area A fitted with two frictionless pistons. The pistons are connected to each other by a metallic wire. Initially, the temperature of the gas is T0 and its pressure is p0 which equals the atmospheric pressure.
a) What is the tension in the wire?
(b) What will be the tension if the temperature is increased to 2T0?
Figure
Ans :
`` \begin{array}{l}\\ \left(\text{a}\right)\\ \text{Since pressure from outside and inside the cylinder is the same, there is no net pressure}\\ \text{acting on the pistons. So, tension will be zero}\text{.}\\ \left(\text{b}\right)\\ {T}_{1}={T}_{o}\\ {T}_{2}=2{T}_{o}\\ {P}_{1}={p}_{o}={10}^{5}\text{Pa}\\ \text{CSA}=A\\ \text{Let the pistons be L distance apart}\text{.}\\ V=AL\\ \text{Applying five variable gas equation, we get}\\ \frac{{P}_{1}V}{{T}_{1}}=\frac{{P}_{2}V}{{T}_{2}}\\ \Rightarrow \frac{{10}^{5}}{{T}_{0}}=\frac{{P}_{2}}{2{T}_{o}}\\ \Rightarrow {P}_{2}=2\times {10}^{5}=2{P}_{o}\\ \text{Net force acting outside = 2}{P}_{0}-{P}_{0}={P}_{0}\\ {\text{Force acting on a piston F= P}}_{o}A\\ \text{By the free body diagram, we get}\\ \mathit{\text{F-T}}\text{=0}\\ {\mathit{\text{T = P}}}_{\mathit{o}}A\end{array}``
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- Qstn #42Figure shows a large closed cylindrical tank containing water. Initially the air trapped above the water surface has a height h0 and pressure 2p0 where p0 is the atmospheric pressure. There is a hole in the wall of the tank at a depth h1 below the top from which water comes out. A long vertical tube is connected as shown.
a) Find the height h2 of the water in the long tube above the top initially.
(b) Find the speed with which water comes out of the hole.
(c) Find the height of the water in the long tube above the top when the water stops coming out of the hole.
Figure
Ans : `` \begin{array}{l}\left(\text{a}\right)\text{Pressure of water above the water level of the bigger tank is given by}\\ P\text{}=({h}_{2}+{h}_{o})\rho g\\ {\text{Let the atmospheric pressure above the tube be P}}_{o}\text{.}\\ \text{Total pressure above the tube =}{P}_{0}+P\\ =({h}_{2}+{h}_{o})\rho g+{P}_{o}\\ {\text{This pressure initially is balanced by pressure above the tank 2P}}_{o}\text{.}\\ \Rightarrow 2{P}_{o}=({h}_{2}+{h}_{o})\rho g+{P}_{o}\\ \Rightarrow {h}_{2}=\frac{{P}_{o}}{\rho g}-{h}_{o}\\ \left(\text{b}\right)\text{Velocity of the efflux out of the outlet depends upon the total pressure above the outlet.}\\ {\text{Total pressure above the outlet = 2P}}_{o}+({h}_{1}-{h}_{o})\rho g\\ \\ \text{Applying Bernouli's law, we get}\\ {\text{Let the velocity of efflux be v}}_{1}{\text{and the velocity with which the level of the tank falls be v}}_{2}.{\text{ Pressure above the outlet is P}}_{o}\text{. Then,}\\ \\ \frac{{\text{2P}}_{o}+({h}_{1}-{h}_{o})\rho g}{\rho }+gz+\frac{{v}_{2}^{2}}{2}=\frac{{\text{P}}_{o}}{\rho }+gz+\frac{{v}_{2}^{2}}{2}\\ \text{Now, let the reference point of the liquid be the level of the outlet. Thus,}\\ \text{z =0}\\ \Rightarrow \frac{{\text{P}}_{o}+({h}_{1}-{h}_{o})\rho g}{\rho }+\frac{{v}_{2}^{2}}{2}=\frac{{v}_{1}^{2}}{2}\\ \text{Again, the speed with which the water level of the tank goes down is very less}\,\mathrm{\,compared\,}\,\mathrm{\,to\,}\,\mathrm{\,the\,}\,\mathrm{\,velocity\,}\,\mathrm{\,of\,}\,\mathrm{\,the\,}\,\mathrm{\,efflux\,}.\,\mathrm{\,Thus\,},\\ {\text{v}}_{2}=0\\ \Rightarrow \frac{{\text{P}}_{o}+({h}_{1}-{h}_{o})\rho g}{\rho }=\frac{{v}_{1}^{2}}{2}\\ \Rightarrow {v}_{1}={\left[\frac{2}{\rho }({\text{P}}_{o}+({h}_{1}-{h}_{o})\rho g)\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\end{array}``
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`` \begin{array}{l}\text{c)}\\ \text{Water maintains its own level, so height of the water of the tank}\\ {\text{will be h}}_{1}\text{}\text{when water will stop flowing}\\ \text{Thus height of water in the tube below the tank height will be =}{h}_{1}\\ \text{Hence height of the water above the tank height will be =}-{h}_{1}\end{array}``
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- Qstn #43An ideal gas is kept in a long cylindrical vessel fitted with a frictionless piston of cross-sectional area 10 cm2 and weight 1 kg (figure). The vessel itself is kept in a big chamber containing air at atmospheric pressure 100 kPa. The length of the gas column is 20 cm. If the chamber is now completely evacuated by an exhaust pump, what will be the length of the gas column? Assume the temperature to remain constant throughout the process.
Figure
Ans : `` \begin{array}{l}\\ \text{Atmospheric pressure}\,\mathrm{\,inside\; the\; cylinderical\,}\,\mathrm{\,vessel\,},{P}_{0}={10}^{5}\text{Pa}\\ {\text{A=10cm}}^{2}=10\times {10}^{-4}{\text{m}}^{2}\\ \text{Pressure due to the weight of the piston =}\frac{mg}{A}=\frac{1\times 9.8}{10\times {10}^{-4}}\\ {P}_{1}={10}^{5}+9.8\times {10}^{3}\\ {V}_{1}=0.2\times 10\times {10}^{-4}=2\times {10}^{-4}\\ \text{After evacution, external pressure above the piston =0}\\ {P}_{2}=0+9.8\times {10}^{3}\\ \text{Now,}\\ {P}_{1}{V}_{1}={P}_{2}{V}_{2}\\ \text{Let L be the final length of the gas column. Then,}\\ {V}_{2}\text{=}10\times {10}^{-4}\text{L}\\ \Rightarrow ({10}^{5}+9.8\times {10}^{3})\times 0.2\times 10\times {10}^{-4}=9.8\times {10}^{3}\times 10\times {10}^{-4}\text{L}\\ \text{⇒L=2}\text{.2 m}\end{array}``
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