Masterclass: 25 Solved JEE Advanced Numericals on Gaseous State
Conquer the kinetic theory of gases, Van der Waals real gas deviations, Graham's Law of effusion, and complex bulb problems. Click "View Solution" to reveal the step-by-step mathematical breakdown.
The States of Matter chapter requires exceptional handling of units (atm vs bar, Joules vs L-atm) and a deep conceptual understanding of when a gas stops behaving ideally. Ensure your grasp of $PV=nRT$ and Kinetic Theory is solid before tackling these!
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Step 1: Understand Open Vessel Conditions
In an open vessel, pressure ($P$) remains constant (atmospheric pressure) and volume ($V$) of the vessel is constant. Therefore, according to the Ideal Gas Law ($PV = nRT$), $n \propto 1/T$, or $n_1 T_1 = n_2 T_2$.
Step 2: Identify Variables
Initial Temperature $T_1 = 27 + 273 = 300 \text{ K}$.
Let initial moles be $n_1 = n$.
Since $3/8$th of the air is expelled, the moles remaining in the vessel are $n_2 = n - \frac{3n}{8} = \frac{5n}{8}$.
Step 3: Solve for Final Temperature ($T_2$)
$n \times 300 = \left( \frac{5n}{8} \right) \times T_2$
$300 = \frac{5}{8} T_2$
$T_2 = \frac{300 \times 8}{5} = 60 \times 8 = 480 \text{ K}$.
Step 4: Convert to Celsius
$T_2 \text{ in }^{\circ}\text{C} = 480 - 273 = 207^{\circ}\text{C}$.
View Solution
Step 1: Conservation of Moles
When connected, the total number of moles remains constant: $n_{\text{total}} = n_A + n_B$.
Using $n = \frac{PV}{RT}$, we get:
$\frac{P_f V_f}{RT} = \frac{P_A V_A}{RT} + \frac{P_B V_B}{RT}$
Step 2: Simplify the Equation
Since $R$ and $T$ are constant, they cancel out:
$P_f V_f = P_A V_A + P_B V_B$
Step 3: Substitute Values
Final volume $V_f = V_A + V_B = 2.0 + 3.0 = 5.0 \text{ L}$.
$P_f (5.0) = (1.0 \times 2.0) + (2.0 \times 3.0)$
$5.0 P_f = 2.0 + 6.0 = 8.0$
$P_f = \frac{8.0}{5.0} = 1.6 \text{ atm}$.
View Solution
Step 1: Relate Rate of Effusion to Time
Rate of effusion ($r$) is Volume effused / Time taken ($r = V/t$).
According to Graham's Law: $\frac{r_A}{r_{O_2}} = \sqrt{\frac{M_{O_2}}{M_A}}$
Step 2: Substitute Volume and Time
Since equal volumes ($50 \text{ mL}$) effuse:
$\frac{V/t_A}{V/t_{O_2}} = \frac{t_{O_2}}{t_A} = \sqrt{\frac{M_{O_2}}{M_A}}$
Step 3: Calculation
$\frac{200}{150} = \sqrt{\frac{32}{M_A}}$
$\frac{4}{3} = \sqrt{\frac{32}{M_A}}$
Square both sides:
$\frac{16}{9} = \frac{32}{M_A}$
$16 \times M_A = 32 \times 9$
$M_A = \frac{288}{16} = 18 \text{ g/mol}$.
View Solution
Step 1: Ideal Gas Density Formula
From $PV = nRT$, substitute $n = m/M$ (mass/Molar Mass).
$PV = \frac{m}{M} RT \implies P \cdot M = \frac{m}{V} RT \implies PM = dRT$
Where $d$ is density.
Step 2: Solve for Molar Mass ($M$)
$M = \frac{dRT}{P}$
$d = 1.964 \text{ g/L}$, $R = 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1}$, $T = 273 \text{ K}$, $P = 1 \text{ atm}$.
$M = \frac{1.964 \times 0.0821 \times 273}{1} \approx 44.0 \text{ g/mol}$.
Step 3: Identify the Alkane
General formula for alkane is ${C_n H_{2n+2}}$.
Molar mass $= 12n + 1(2n + 2) = 14n + 2$.
$14n + 2 = 44 \implies 14n = 42 \implies n = 3$.
The alkane is ${C_3H_8}$ (Propane).
View Solution
Step 1: Apply Dalton's Law for Dry Gas Pressure
$P_{\text{total}} = P_{\text{dry gas}} + P_{\text{water vapor (aqueous tension)}}$
$P_{\text{dry } N_2} = 750 - 25 = 725 \text{ mm Hg}$.
Step 2: Setup Combined Gas Law
Initial state (Dry Gas): $P_1 = 725 \text{ mm Hg}$, $V_1 = 400 \text{ mL}$, $T_1 = 27 + 273 = 300 \text{ K}$.
Final state (STP): $P_2 = 760 \text{ mm Hg}$, $T_2 = 273 \text{ K}$, $V_2 = ?$
Step 3: Calculate $V_2$
$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$
$\frac{725 \times 400}{300} = \frac{760 \times V_2}{273}$
$V_2 = \frac{725 \times 400 \times 273}{300 \times 760}$
$V_2 = \frac{79170000}{228000} \approx 347.2 \text{ mL}$.
View Solution
Step 1: Calculate Volume of Balloon
$V = \frac{4}{3} \pi r^3 = \frac{4}{3} \times 3.1416 \times 10^3 \approx 4188.8 \text{ m}^3 = 4.1888 \times 10^6 \text{ L}$.
Step 2: Calculate Mass of Helium Inside ($m_{He}$)
From $PV = \frac{m}{M} RT \implies m_{He} = \frac{PVM}{RT}$
$m_{He} = \frac{1 \times 4.1888 \times 10^6 \times 4}{0.082 \times 300} = \frac{1.6755 \times 10^7}{24.6}$
$m_{He} \approx 681,100 \text{ g} = 681.1 \text{ kg}$.
Step 3: Calculate Mass of Displaced Air
Mass of air displaced $= \text{Volume} \times \text{Density of air}$
Mass of air $= 4188.8 \text{ m}^3 \times 1.2 \text{ kg/m}^3 \approx 5026.6 \text{ kg}$.
Step 4: Calculate Payload
Payload $= \text{Mass of displaced air} - (\text{Mass of balloon} + \text{Mass of He})$
Payload $= 5026.6 - (100 + 681.1) = 5026.6 - 781.1 = 4245.5 \text{ kg}$.
View Solution
Step 1: Formula for Kinetic Energy
The total translational kinetic energy for $n$ moles of an ideal gas is $KE = \frac{3}{2} nRT$.
Step 2: Determine Moles ($n$)
Molar mass of ${N_2} = 28 \text{ g/mol}$.
Moles $n = \frac{14 \text{ g}}{28 \text{ g/mol}} = 0.5 \text{ mol}$.
Step 3: Calculation
$T = 127 + 273 = 400 \text{ K}$.
$KE = \frac{3}{2} \times 0.5 \times 8.314 \times 400$
$KE = 1.5 \times 0.5 \times 3325.6$
$KE = 0.75 \times 3325.6 = 2494.2 \text{ Joules}$.
View Solution
Step 1: Formula for RMS Velocity
$V_{\text{rms}} = \sqrt{\frac{3RT}{M}}$
Step 2: Equate the Velocities
$V_{\text{rms}} ({O_2}) = V_{\text{rms}} ({H_2})$
$\sqrt{\frac{3 R T_{O_2}}{M_{O_2}}} = \sqrt{\frac{3 R T_{H_2}}{M_{H_2}}}$
Step 3: Simplify and Solve
Square both sides and cancel $3R$:
$\frac{T_{O_2}}{M_{O_2}} = \frac{T_{H_2}}{M_{H_2}}$
Substitute $M_{O_2} = 32$, $M_{H_2} = 2$, $T_{H_2} = 20 \text{ K}$:
$\frac{T_{O_2}}{32} = \frac{20}{2}$
$\frac{T_{O_2}}{32} = 10 \implies T_{O_2} = 320 \text{ K}$.
View Solution
Step 1: Convert Weight to Moles
Let the total mass of the mixture be $100 \text{ g}$.
Mass of ${H_2} = 20 \text{ g}$. Moles of ${H_2} = 20 / 2 = 10 \text{ moles}$.
Mass of ${O_2} = 80 \text{ g}$. Moles of ${O_2} = 80 / 32 = 2.5 \text{ moles}$.
Step 2: Calculate Mole Fraction of Hydrogen ($x_{H_2}$)
Total moles $= 10 + 2.5 = 12.5 \text{ moles}$.
$x_{H_2} = \frac{10}{12.5} = \frac{100}{125} = 0.8$.
Step 3: Apply Dalton's Law
$P_{H_2} = x_{H_2} \times P_{\text{total}}$
$P_{H_2} = 0.8 \times 3.0 = 2.4 \text{ atm}$.
View Solution
Step 1: Use Compressibility Equation
$V = \frac{Z \cdot nRT}{P}$
Step 2: Substitute Values
$V = \frac{0.8 \times 1 \times 0.0821 \times 273}{100}$
$V = \frac{0.8 \times 22.41}{100} = \frac{17.93}{100} = 0.1793 \text{ L} = 179.3 \text{ mL}$.
Step 3: Analyze Z value
Since $Z < 1$, the real gas occupies less volume than an ideal gas would under the same conditions. This means attractive forces dominate, making the gas more compressible than an ideal gas.
View Solution
Step 1: Write Van der Waals Equation
$\left(P + \frac{an^2}{V^2}\right)(V - nb) = nRT$
Step 2: Rearrange for Pressure ($P$)
$P = \frac{nRT}{V - nb} - \frac{an^2}{V^2}$
Step 3: Substitute Values
$n = 1$, $V = 0.5$, $T = 273$, $R = 0.0821$.
$P = \frac{1 \times 0.0821 \times 273}{0.5 - (1 \times 0.0427)} - \frac{3.59 \times 1^2}{0.5^2}$
$P = \frac{22.41}{0.5 - 0.0427} - \frac{3.59}{0.25}$
$P = \frac{22.41}{0.4573} - 14.36$
$P = 49.0 - 14.36 = 34.64 \text{ atm}$.
Note: For an ideal gas, $P = 22.41 / 0.5 = 44.82 \text{ atm}$. The real pressure is much lower due to the strong attractive forces '$a$'.
View Solution
Step 1: Relate Distance to Rate
Since time ($t$) is the same for both gases, distance traveled ($x$) is proportional to their rate of effusion ($r$).
$\frac{x_{NH_3}}{x_{HCl}} = \frac{r_{NH_3}}{r_{HCl}} = \sqrt{\frac{M_{HCl}}{M_{NH_3}}}$
Step 2: Setup Molar Masses and Distances
$M_{NH_3} = 17 \text{ g/mol}$, $M_{HCl} = 36.5 \text{ g/mol}$.
Let distance traveled by $NH_3$ be $x$. The distance traveled by $HCl$ will be $(100 - x)$.
Step 3: Solve the Equation
$\frac{x}{100 - x} = \sqrt{\frac{36.5}{17}} = \sqrt{2.147} \approx 1.465$
$x = 1.465 (100 - x)$
$x = 146.5 - 1.465x$
$2.465x = 146.5$
$x = \frac{146.5}{2.465} \approx 59.4 \text{ cm}$.
View Solution
Step 1: Formulas for Critical Constants
$P_c = \frac{a}{27b^2}$
$T_c = \frac{8a}{27Rb}$
Step 2: Find constant '$a$'
From $P_c$: $a = 27 \cdot b^2 \cdot P_c$
$a = 27 \times (0.04)^2 \times 50$
$a = 27 \times 0.0016 \times 50 = 27 \times 0.08 = 2.16 \text{ L}^2\text{ atm mol}^{-2}$.
Step 3: Calculate $T_c$
$T_c = \frac{8 \times 2.16}{27 \times 0.0821 \times 0.04}$
$T_c = \frac{17.28}{0.088668} \approx 194.9 \text{ K}$.
Alternative shortcut: $\frac{T_c}{P_c} = \frac{8b}{R} \implies T_c = P_c \frac{8b}{R} = 50 \times \frac{8 \times 0.04}{0.0821} = \frac{16}{0.0821} \approx 194.9 \text{ K}$.
View Solution
Step 1: Modified Graham's Law for Mixtures
When a mixture effuses, the rate of effusion of each component is directly proportional to its partial pressure (or initial mole fraction) and inversely proportional to the square root of its molar mass.
$\frac{n'_{He}}{n'_{CH_4}} = \frac{n_{He}}{n_{CH_4}} \times \sqrt{\frac{M_{CH_4}}{M_{He}}}$
Where $n'$ represents the moles effused.
Step 2: Substitute Values
Initial molar ratio $\frac{n_{He}}{n_{CH_4}} = \frac{1}{1}$.
$M_{He} = 4 \text{ g/mol}$, $M_{CH_4} = 16 \text{ g/mol}$.
$\frac{n'_{He}}{n'_{CH_4}} = 1 \times \sqrt{\frac{16}{4}} = \sqrt{4} = 2$.
View Solution
Step 1: Write the Reaction
${CaCO_3}_{(s)} \xrightarrow{\Delta} CaO_{(s)} + {CO_2}_{(g)}$
Note that only ${CO_2}$ gas exerts pressure. Solids do not contribute to gas pressure.
Step 2: Calculate Moles of ${CO_2}$
Moles of ${CaCO_3} = \frac{20 \text{ g}}{100 \text{ g/mol}} = 0.2 \text{ moles}$.
From stoichiometry, 1 mole of ${CaCO_3}$ produces 1 mole of ${CO_2}$.
Moles of ${CO_2} (n) = 0.2 \text{ moles}$.
Step 3: Apply Ideal Gas Law
$P = \frac{nRT}{V} = \frac{0.2 \times 0.082 \times 800}{2.0}$
$P = \frac{0.2 \times 65.6}{2.0} = \frac{13.12}{2.0} = 6.56 \text{ atm}$.
View Solution
Step 1: Formula for Boyle Temperature
The Boyle temperature is mathematically related to the Van der Waals constants by the formula:
$T_B = \frac{a}{Rb}$
Step 2: Substitute Values
$R = 0.0821 \text{ L atm K}^{-1} \text{ mol}^{-1}$.
$T_B = \frac{1.39}{0.0821 \times 0.0391}$
Step 3: Calculate
$T_B = \frac{1.39}{0.00321} \approx 433 \text{ K}$.
View Solution
Step 1: Understand the Faulty Barometer
True Pressure ($P_{\text{true}}$) = Barometer Reading ($h$) + Pressure of trapped air ($P_{\text{air}}$).
Initial state: $760 = 740 + P_{\text{air}} \implies P_{\text{air1}} = 20 \text{ mm Hg}$.
Volume of trapped air $V_1 \propto \text{length} = 10 \text{ cm}$.
Step 2: Second State Setup
New reading $h_2 = 750 \text{ mm Hg}$.
Because the mercury column rose by $10 \text{ mm}$ ($1 \text{ cm}$), the trapped air was compressed.
New length of air column $L_2 = 10 \text{ cm} - 1 \text{ cm} = 9 \text{ cm}$. So $V_2 \propto 9 \text{ cm}$.
Step 3: Apply Boyle's Law to Trapped Air
$P_{\text{air1}} V_1 = P_{\text{air2}} V_2$
$20 \times 10 = P_{\text{air2}} \times 9$
$P_{\text{air2}} = \frac{200}{9} \approx 22.22 \text{ mm Hg}$.
Step 4: Calculate New True Pressure
$P_{\text{true2}} = h_2 + P_{\text{air2}} = 750 + 22.22 = 772.22 \text{ mm Hg}$.
View Solution
Step 1: Calculate Average Molar Mass of Mixture ($M_{\text{mix}}$)
$M_{\text{mix}} = \frac{dRT}{P} = \frac{2.33 \times 0.0821 \times 300}{1} \approx 57.4 \text{ g/mol}$.
Step 2: Formula linking Molar Mass and $\alpha$
Theoretical molar mass of undecomposed ${N_2O_4}$ ($M_{\text{theoretical}}$) $= 2(14) + 4(16) = 92 \text{ g/mol}$.
For reaction $A \rightleftharpoons nB$, $\alpha = \frac{M_{\text{theoretical}} - M_{\text{mix}}}{(n-1) M_{\text{mix}}}$.
Step 3: Calculate $\alpha$
Here, 1 mole of ${N_2O_4}$ gives 2 moles of ${NO_2}$, so $n = 2$.
$\alpha = \frac{92 - 57.4}{(2-1) \times 57.4} = \frac{34.6}{57.4} \approx 0.603$.
View Solution
Step 1: Analyze High Pressure Condition
Van der Waals equation for 1 mole: $\left(P + \frac{a}{V^2}\right)(V - b) = RT$.
At extremely high pressure, the volume $V$ becomes very small. The physical volume of the molecules ($b$) cannot be ignored compared to $V$. However, the pressure $P$ is so massive that the internal attraction correction term ($\frac{a}{V^2}$) becomes negligible compared to $P$.
Step 2: Simplify Equation
Assume $P + \frac{a}{V^2} \approx P$.
$P(V - b) = RT \implies PV - Pb = RT$.
Step 3: Derive Z
Divide the entire equation by $RT$:
$\frac{PV}{RT} - \frac{Pb}{RT} = 1$
Since $Z = \frac{PV}{RT}$, we get:
$Z - \frac{Pb}{RT} = 1 \implies Z = 1 + \frac{Pb}{RT}$.
View Solution
Step 1: Formula for Mean Free Path
$\lambda = \frac{1}{\sqrt{2} \pi \sigma^2 n^*}$
Where $\sigma$ is the collision diameter and $n^*$ is the number density (molecules per unit volume, $N/V$).
Step 2: Relate Number Density to Pressure
From Ideal Gas Law: $PV = N k_B T$ (using Boltzmann constant $k_B$).
$\frac{N}{V} = n^* = \frac{P}{k_B T}$.
Step 3: Substitute and Analyze
$\lambda = \frac{k_B T}{\sqrt{2} \pi \sigma^2 P}$.
This shows $\lambda$ is inversely proportional to Pressure ($P$) at constant temperature.
Step 4: Conclusion
If pressure is doubled ($P \rightarrow 2P$), the mean free path is halved.
View Solution
Step 1: Formula for Isothermal Reversible Expansion
$W = -nRT \ln \left(\frac{V_2}{V_1}\right) = -2.303 nRT \log_{10} \left(\frac{V_2}{V_1}\right)$
Step 2: Substitute Values
$W = -2.303 \times 1 \times 8.314 \times 300 \times \log_{10} \left(\frac{20.0}{2.0}\right)$
$W = -2.303 \times 2494.2 \times \log_{10} (10)$
Since $\log_{10}(10) = 1$:
Step 3: Calculate
$W = -5744.1 \text{ Joules}$.
View Solution
Step 1: Formula
Average kinetic energy per mole is $\frac{3}{2}RT$.
Average kinetic energy per molecule is $\frac{3}{2}k_B T$.
Step 2: Substitute Values
$T = 27 + 273 = 300 \text{ K}$.
$KE = \frac{3}{2} \times (1.38 \times 10^{-23}) \times 300$
$KE = 1.5 \times 1.38 \times 10^{-23} \times 300$
$KE = 450 \times 1.38 \times 10^{-23} = 621 \times 10^{-23}$.
Step 3: Scientific Notation
$KE = 6.21 \times 10^{-21} \text{ Joules}$.
View Solution
Step 1: Expansion of Van der Waals Equation
By algebraically expanding the Van der Waals equation into the Virial form, the coefficient $B$ matches the $1/V_m$ term.
Derivation yields: $B = b - \frac{a}{RT}$.
Step 2: Analyze Boyle Temperature Condition
At the Boyle temperature ($T_B$), the gas behaves ideally over a wide range of pressures, which mathematically means the second virial coefficient $B$ must equal zero.
Step 3: Verification
Set $B = 0 \implies b - \frac{a}{RT_B} = 0 \implies b = \frac{a}{RT_B} \implies T_B = \frac{a}{Rb}$. This matches our previous formula perfectly!
View Solution
Step 1: Identify Formulas
$V_{mp} = \sqrt{\frac{2RT}{M}}$
$V_{rms} = \sqrt{\frac{3RT}{M}}$
Step 2: Establish the Ratio
$\frac{V_{rms}}{V_{mp}} = \frac{\sqrt{3RT/M}}{\sqrt{2RT/M}} = \sqrt{\frac{3}{2}} \approx 1.224$.
Step 3: Calculate
$V_{rms} = V_{mp} \times \sqrt{1.5} = 400 \times 1.224 = 489.6 \text{ m/s}$.
View Solution
Step 1: Analyze Volume Changes
$KOH$ absorbs ${CO_2}$. Reduction by $KOH = 70 - 30 = 40 \text{ mL}$. Thus, Volume of ${CO_2}$ formed $= 40 \text{ mL}$.
Residual gas after $KOH$ is unreacted Oxygen. Unreacted ${O_2} = 30 \text{ mL}$.
Oxygen consumed $= \text{Initial} - \text{Unreacted} = 100 - 30 = 70 \text{ mL}$.
Step 2: Use General Combustion Equation
$C_xH_y + (x + y/4)O_2 \rightarrow xCO_2 + (y/2)H_2O_{(l)}$
Step 3: Solve for x (Carbon)
$1 \text{ vol}$ hydrocarbon gives $x \text{ vol } CO_2$.
$10 \text{ mL} \times x = 40 \text{ mL} \implies x = 4$.
Step 4: Solve for y (Hydrogen)
$1 \text{ vol}$ hydrocarbon consumes $(x + y/4) \text{ vol } O_2$.
$10 \times (4 + y/4) = 70$
$4 + y/4 = 7 \implies y/4 = 3 \implies y = 12$.
Ah! $10 \times (x + y/4) = 70$. Since $x=4$, $4 + y/4 = 7 \implies y/4 = 3 \implies y = 12$.
The math implies $C_4H_{12}$, which doesn't exist chemically! A classic competitive exam trick: The initial data provided in this specific question ($70\text{mL}$ residual) is chemically inconsistent for a standard stable hydrocarbon. If residual was $85\text{mL}$, $O_2$ used $= 55\text{mL}$, yielding $C_4H_6$. This teaches you to trust your algebra but verify chemical reality!
Mastering the Gas Laws
Congratulations on powering through these 25 complex numericals. The Gaseous State is the ultimate test of balancing algebra with physical reality. By mastering how to connect microscopic kinetic parameters (like RMS speed and collision theory) with macroscopic thermodynamic variables ($P, V, T, Z$), you have laid the groundwork for tackling the hardest problems in physical chemistry. Keep an eye out for units, specifically when using $R$ in Joules ($8.314$) versus L-atm ($0.0821$)!
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