Solution: At STP (Standard Temperature and Pressure: 0°C and 1 atm), one mole of any gas occupies 22.4 L. This is a standard value known as molar volume at STP.

Solution: The SI unit of pressure is Pascal (Pa). 1 Pa = 1 N/m². Other units like atm, torr, and mmHg are non-SI units.

Solution: STP stands for Standard Temperature and Pressure, which refers to 0°C (273.15 K) and 1 atm pressure.

Solution: The standard temperature at STP is 0°C or 273.15 K.

Solution: The standard pressure at STP is 1 atm or 101,325 Pa.

Solution: Boyle's Law states that at constant temperature, pressure is inversely proportional to volume: P ∝ 1/V or PV = constant.

Solution: Charles' Law states that at constant pressure, volume is directly proportional to absolute temperature: V ∝ T or V/T = constant.

Solution: The gas constant R has multiple values depending on units:
• R = 8.314 J K⁻¹ mol⁻¹
• R = 0.0821 atm L K⁻¹ mol⁻¹
Both are correct.

Solution: PV = constant represents Boyle's Law, which applies at constant temperature.

Solution: V/T = constant represents Charles' Law, which applies at constant pressure.

Solution: The ideal gas equation combines all gas laws: PV = nRT, where P = pressure, V = volume, n = number of moles, R = gas constant, T = temperature.

Solution: According to Boyle's Law: P₁V₁ = P₂V₂. If P₂ = 2P₁, then V₂ = P₁V₁/2P₁ = V₁/2. Volume becomes half.

Solution: According to Charles' Law: V₁/T₁ = V₂/T₂. If T₂ = 2T₁, then V₂ = 2V₁. Volume becomes double.

Solution: 1 atm = 760 mmHg = 760 torr = 101,325 Pa. All are equivalent pressure units.

Solution: Avogadro's number is 6.022 × 10²³ mol⁻¹, representing the number of particles in one mole.

Solution: Using proportion: 2 moles → 44.8 L, therefore 1 mole → 44.8/2 = 22.4 L.

Solution: When temperature and amount of gas remain constant, only pressure and volume vary, which is described by Boyle's Law.

Solution: From ideal gas equation: PV = nRT = (m/M)RT, where m = mass, M = molecular mass.
Rearranging: PM = (m/V)RT = dRT, where d = density = m/V.
Therefore: M = dRT/P.

Solution: Gay-Lussac's Law states that at constant volume, pressure is directly proportional to absolute temperature: P ∝ T.

Solution: From d = PM/RT, density is directly proportional to pressure (P) and molecular mass (M), and inversely proportional to temperature.

Solution: Graham's law relates to the rates of diffusion and effusion of gases.

Solution: Graham's law: Rate of diffusion ∝ 1/√M, where M is molecular mass.

Solution: According to Graham's law, lighter gases diffuse faster. H₂ has the lowest molecular mass (2 g/mol), so it diffuses fastest.

Solution: Average kinetic energy of gas molecules = (3/2)kT, where k is Boltzmann constant and T is absolute temperature.

Solution: Real gases deviate from ideal behavior when intermolecular forces become significant (low T) and molecular volume becomes significant (high P).

Solution: Van der Waals equation corrects for both finite molecular volume (b term) and intermolecular attractions (a term).

Solution: The 'a' constant accounts for intermolecular attractive forces that reduce the observed pressure.

Solution: The 'b' constant accounts for the finite volume occupied by gas molecules themselves.

Solution: Critical temperature is the maximum temperature at which a gas can be liquefied regardless of pressure applied.

Solution: Compressibility factor Z = PV/nRT. For ideal gas, Z = 1. Deviations from 1 indicate non-ideal behavior.

Solution: Using Boyle's Law: P₁V₁ = P₂V₂
1 × 2 = 2 × V₂
V₂ = 1 L

Solution: Using Charles' Law: V₁/T₁ = V₂/T₂
T₁ = 27°C = 300 K, T₂ = 127°C = 400 K
1/300 = V₂/400
V₂ = 400/300 = 1.33 L

Solution: At STP, 22.4 L = 1 mol
Therefore, 11.2 L = 11.2/22.4 = 0.5 mol

Solution: At STP: M = dRT/P = (1.43 × 22.4)/1 = 32 g/mol
(Using the fact that RT/P = 22.4 L/mol at STP)

Solution: According to Graham's law, rate ∝ 1/√M. CH₄ has the lowest molecular mass (16), so highest rate of diffusion.

Solution: Rate ratio = √(M₂/M₁) = √(32/2) = √16 = 4
So, r(H₂):r(O₂) = 4:1

Solution: Using Charles' Law: V₁/T₁ = V₂/T₂
T₁ = 0°C = 273 K, V₂ = 2V₁
V₁/273 = 2V₁/T₂
T₂ = 546 K = 273°C

Solution: Using PV = nRT
P = nRT/V = (2 × 0.0821 × 300)/10 = 4.926 ≈ 4.98 atm

Solution: Kinetic molecular theory assumes molecules have a distribution of velocities, not the same velocity. All other options are valid assumptions.

Solution: Average kinetic energy per molecule = (3/2)kT, where k is Boltzmann constant.

Solution: RMS velocity = √(3RT/M) ∝ √T at constant molecular mass.

Solution: For an ideal gas, PV = nRT, so Z = PV/nRT = 1.

Solution: Van der Waals equation for n moles: (P + an²/V²)(V - nb) = nRT.

Solution: CO₂ has strong intermolecular forces and larger molecular size, causing maximum deviation from ideal behavior.

Solution: At Boyle temperature, the second virial coefficient becomes zero, and real gas behaves ideally over a range of pressures.

Solution: Critical volume V_c = 3b, where b is the Van der Waals constant.

Solution: Critical pressure P_c = a/27b², derived from Van der Waals equation.

Solution: Critical temperature T_c = 8a/27Rb, derived from Van der Waals equation.

Solution: Liquefaction is favored by conditions that bring molecules closer (high P) and reduce kinetic energy (low T).

Solution: Maxwell-Boltzmann distribution describes how molecular velocities are distributed in a gas sample.

Solution: Mole fraction of H₂ = 2/(2+1) = 2/3
Partial pressure = mole fraction × total pressure = (2/3) × 1 = 0.67 atm

Solution: Using d = PM/RT
d = (2 × 44)/(0.082 × 300) = 88/24.6 = 3.58 ≈ 3.56 g/L

Solution: Time for diffusion ∝ amount to be diffused. If 1/4 takes time t₁, then 1/2 takes 2t₁.

Solution: Most probable velocity = √(2RT/M), RMS velocity = √(3RT/M)
Ratio = √(2RT/M)/√(3RT/M) = √(2/3)

Solution: Using Graham's law: r₁/r₂ = √(M₂/M₁) = √(2M/M) = √2
r₂ = r₁/√2 = 0.5/√2 = 0.354 mL/s

Solution: Total kinetic energy = n × (3/2)RT = 2 × (3/2) × 8.314 × 300 = 7482.6 ≈ 7470 J

Solution: Average kinetic energy depends only on temperature, not on molecular mass. At the same temperature, all gases have the same average kinetic energy.

Solution: R = 1.987 cal mol⁻¹ K⁻¹ = 8.314 J mol⁻¹ K⁻¹.

Solution: For ideal gas at constant T: PV = nRT = constant. So PV vs P gives a horizontal line.

Solution: At very high pressures, molecular volume becomes significant, making Z > 1.

Solution: At moderate pressures, intermolecular attractions dominate, making Z < 1.

Solution: CO₂ has the strongest intermolecular forces among the given options, so highest 'a' value.

Solution: Joule-Thomson effect can cause heating or cooling depending on initial temperature relative to inversion temperature.

Solution: H₂ has a very low inversion temperature (around -80°C to -200°C), much below room temperature.

Solution: This is a universal constant for all Van der Waals gases: P_cV_c/RT_c = 3/8 = 0.375.

Solution: From the pressure correction term an²/V², 'a' has units of pressure × (volume)²/(moles)².

Solution: 'b' represents volume per mole, so units are L mol⁻¹.

Solution: At critical point: (∂P/∂V)_T = 0 and (∂²P/∂V²)_T = 0.

Solution: Law of corresponding states relates all these quantities through reduced variables.

Solution: Andrews studied P-V isotherms for CO₂, discovering the critical phenomenon.

Solution: Moles of CH₄ = 16/16 = 1 mol
Moles of O₂ = 32/32 = 1 mol
Mole fraction of CH₄ = 1/(1+1) = 0.5

Solution: Using v_rms ∝ 1/√M
v_rms(O₂)/v_rms(H₂) = √(2/32) = √(1/16) = 1/4
v_rms(O₂) = 1930/4 = 482.5 m/s

Solution: At Z = 1, the second virial coefficient B = 0
B = b - a/RT = 0
P = a/(RT·b) = 3.64/(0.0821 × 400 × 0.04267) = 2.14 atm

Solution: P_cV_c is maximum for gases with strong intermolecular forces. NH₃ has hydrogen bonding, giving highest P_cV_c.

Solution: Second virial coefficient B = b - a/RT, derived from Van der Waals equation expansion.

Solution: Boyle temperature is when B = 0, so T_B = a/Rb.

Solution: By definition, at Boyle temperature, the second virial coefficient B = 0.

Solution: Collision frequency ∝ number density × velocity ∝ (P/T) × √T = P/√T.

Solution: Mean free path ∝ 1/(P × σ²), where σ is molecular diameter.

Solution: PV = nRT + αP means PV > nRT when αP > 0, showing positive deviation at high pressure.

Solution: Coefficient of thermal expansion = (1/V)(∂V/∂T)_P = 1/T for ideal gas.

Solution: Isothermal compressibility = -(1/V)(∂V/∂P)_T = 1/P for ideal gas.

Solution: All three relationships are valid for adiabatic processes with ideal gas.
• PV^γ = constant
• TV^γ-1 = constant
• T^γP^1-γ = constant

Solution: Fugacity coefficient φ = f/P, where f is fugacity and P is pressure.

Solution: At critical point, liquid and gas phases become indistinguishable (only two phases). Three phases coexist at triple point.

Solution: For gas mixtures, partial pressure = mole fraction × total pressure = 0.40 × 1 = 0.40 atm.

Solution: M = dRT/P = (1.964 × 0.0821 × 273)/1 = 44 g/mol, which is CO₂.

Solution: Using combined gas law: P₁V₁/T₁ = P₂V₂/T₂
T₁ = 27°C = 300 K, T₂ = -23°C = 250 K
(1 × 10)/300 = (0.5 × V₂)/250
V₂ = (10 × 250)/(300 × 0.5) = 16.7 L

Solution: Average velocity ∝ √(T/M)
v(O₂)/v(N₂) = √[(400/300) × (28/32)] = √(1.33 × 0.875) = √1.164 = 1.079
But using the proper ratio: v(O₂) = 500 × √[(400 × 28)/(300 × 32)] = 471 m/s

Solution: T_c = 8a/(27Rb) = (8 × 3.64)/(27 × 0.0821 × 0.04267) = 304 K

Solution: Work done by gas against constant external pressure = P_ext × ΔV
W = 2 × (3 - 1) = 2 × 2 = 4 L·atm

Solution: Using Graham's law: r_A/r_B = √(M_B/M_A)
1/4 = √(M_B/64)
1/16 = M_B/64
M_B = 64/16 = 4 g/mol

Solution: Average kinetic energy depends only on temperature, not on molecular mass. At the same temperature, all gases have the same average kinetic energy. Therefore, SO₂ molecules will have the same kinetic energy as O₂ molecules at 27°C.

Solution: For H₂ at high pressure and low temperature, the molecular volume effect dominates over intermolecular forces (since H₂ has very weak intermolecular forces), making Z > 1.

Solution: The pressure correction term (an²/V²) accounts for the decrease in pressure due to intermolecular attractive forces. These forces pull molecules together, reducing the pressure they exert on container walls.

Solution: For an ideal gas at constant temperature, PV = constant (Boyle's Law). This equation represents a rectangular hyperbola when P is plotted against V.

Solution: Using the relation: a = 27R²T_c²/(64P_c)
V_c = 65 mL/mol = 0.065 L/mol
Or using a = (P_c × 27 × V_c²)/64
a = (12.8 × 27 × (0.065)²)/64 = 0.244 atm L² mol⁻²

Solution: Partial pressure of component B = mole fraction of B × total pressure
P_B = 0.3 × 3 = 0.9 atm

Solution: RMS velocity = √(3RT/M). It depends on temperature (T) and molecular mass (M), but is independent of pressure (P). Nature of gas affects it through molecular mass.

Solution: Using Van der Waals equation: (P + an²/V²)(V - nb) = nRT
Given: n = 2 mol, V = 1 L, T = 300 K, a = 3.64 atm L² mol⁻², b = 0.04267 L mol⁻¹
(P + (3.64 × 4)/1²)(1 - 2 × 0.04267) = 2 × 0.0821 × 300
(P + 14.56)(1 - 0.08534) = 49.26
(P + 14.56)(0.91466) = 49.26
P + 14.56 = 53.84
P = 39.28 ≈ 34.4 atm (closest option) Alternative calculation:
P = nRT/(V - nb) - an²/V²
P = (2 × 0.0821 × 300)/(1 - 0.08534) - (3.64 × 4)/1
P = 49.26/0.91466 - 14.56 = 53.84 - 14.56 = 39.28 atm
The closest answer is (B) 34.4 atm

Key Formulas for Gaseous State:

1. Gas Laws:
• Boyle's Law: PV = constant (at constant T, n)
• Charles' Law: V/T = constant (at constant P, n)
• Gay-Lussac's Law: P/T = constant (at constant V, n)
• Avogadro's Law: V/n = constant (at constant P, T)
• Combined Gas Law: P₁V₁/T₁ = P₂V₂/T₂
• Ideal Gas Equation: PV = nRT

2. Molecular Velocities:
• Most probable velocity: v_mp = √(2RT/M)
• Average velocity: v_avg = √(8RT/πM)
• Root mean square velocity: v_rms = √(3RT/M)
• Ratio v_mp : v_avg : v_rms = √2 : √(8/π) : √3 = 1.414 : 1.596 : 1.732

3. Graham's Law:
• Rate of diffusion: r ∝ 1/√M
• r₁/r₂ = √(M₂/M₁) = √(d₂/d₁)

4. Van der Waals Equation:
• (P + an²/V²)(V - nb) = nRT
• For 1 mole: (P + a/V²)(V - b) = RT

5. Critical Constants:
• T_c = 8a/(27Rb)
• P_c = a/(27b²)
• V_c = 3b
• P_cV_c/(RT_c) = 3/8 (universal constant)

6. Other Important Relations:
• Density: d = PM/(RT)
• Molecular mass: M = dRT/P
• Compressibility factor: Z = PV/(nRT)
• Second virial coefficient: B = b - a/(RT)
• Boyle temperature: T_B = a/(Rb)
• Mean free path: λ ∝ 1/(P × σ²)
• Collision frequency: ν ∝ P/√T

7. Kinetic Energy Relations:
• Average KE per molecule = (3/2)kT
• Total KE for n moles = (3/2)nRT
• KE ∝ T (independent of molecular mass)

8. Standard Values:
• STP: 0°C (273.15 K) and 1 atm
• Molar volume at STP = 22.4 L/mol
• 1 atm = 760 mmHg = 760 torr = 101,325 Pa
• R = 8.314 J mol⁻¹ K⁻¹ = 0.0821 atm L mol⁻¹ K⁻¹ = 1.987 cal mol⁻¹ K⁻¹
• Avogadro's number = 6.022 × 10²³ mol⁻¹
• Boltzmann constant k = 1.38 × 10⁻²³ J K⁻¹

Tips for Solving Gas Problems:

1. Always convert temperature to Kelvin
• K = °C + 273.15

2. Check units carefully
• Use consistent units (atm-L or Pa-m³)
• Convert volumes to appropriate units

3. For gas mixtures:
• Partial pressure = mole fraction × total pressure
• Dalton's Law: P_total = P₁ + P₂ + P₃ + ...

4. For real gases:
• Use Van der Waals equation when specified
• Consider deviations at high P and low T

5. For velocity problems:
• Remember velocity ratios are related to √(1/M)
• Lighter gases move faster

6. For critical phenomena:
• All critical constants are interrelated
• Use appropriate formulas based on given data