Gas Laws & Real Substances
1 / 50
A mixture of 2 moles of CO and 3 moles of CO₂ is at 500 K and a total volume of 0.2 m³. The total pressure of the mixture is:
5.195 bar
2.598 bar
20.78 bar
10.39 bar
Total moles n_total = 2 + 3 = 5. Using PV = nRT, P = (nRT)/V = (5 × 8.314 × 500)/0.2 = 20770 Pa = 20770 × 10⁻⁵ bar ≈ 10.39 bar.
2 / 50
The partial pressure of a gas in a mixture is 0.4 bar, and the total pressure is 2 bar. The partial volume of this gas, if the total volume is 0.1 m³, is:
0.02 m³
0.04 m³
0.08 m³
0.05 m³
Mole fraction yᵢ = Pᵢ/P_total = 0.4/2 = 0.2. By Amagat’s law, Vᵢ = yᵢ × V_total = 0.2 × 0.1 = 0.02 m³.
3 / 50
A gas mixture contains 1 kg of H₂ (molar mass 2 kg/kmol) and 2 kg of N₂ (molar mass 28 kg/kmol) at 1 bar and 300 K. The total volume of the mixture is:
172.6 m³
86.3 m³
43.15 m³
344.2 m³
Moles of H₂ = 1/2 = 0.5 kmol. Moles of N₂ = 2/28 = 1/14 ≈ 0.0714 kmol. Total moles n_total = 0.5 + 0.0714 = 0.5714 kmol. Using PV = nRT, V = (nRT)/P = (0.5714 × 8314 × 300)/(1 × 10⁵) ≈ 86.3 m³ (R = 8314 J/kmol·K).
4 / 50
A vessel contains a mixture of 40% N₂ and 60% O₂ by volume at 2 bar and 300 K. The partial pressure of O₂ is:
0.8 bar
1.2 bar
1.0 bar
1.6 bar
By Amagat’s law, volume fraction equals mole fraction for ideal gases at the same P and T. Thus, mole fraction of O₂ = 0.6. By Dalton’s law, P_O₂ = 0.6 × 2 = 1.2 bar.
5 / 50
A gas mixture at 400 K and 3 bar consists of 2 moles of CH₄ and 3 moles of Ar. The partial pressure of Ar is:
1.8 bar
2.4 bar
0.6 bar
Total moles n_total = 2 + 3 = 5. Mole fraction of Ar = 3/5. By Dalton’s law, P_Ar = (3/5) × 3 = 1.8 bar.
6 / 50
The mole fraction of a gas in a mixture is equal to:
The ratio of its partial pressure to the total pressure
The ratio of its partial volume to the total volume
Both A and B
The ratio of its mass to the total mass
For an ideal gas mixture, the mole fraction yᵢ = nᵢ/n_total. By Dalton’s law, yᵢ = Pᵢ/P_total (since Pᵢ = (nᵢ RT/V) and P_total = (n_total RT/V)). By Amagat’s law, yᵢ = Vᵢ/V_total (since Vᵢ = (nᵢ RT/P) and V_total = (n_total RT/P)). Thus, both A and B are correct.
7 / 50
A mixture of 1 mole of CO₂ and 4 moles of He is at 1 bar and 300 K. The partial volume of CO₂ according to Amagat’s law is:
0.00498 m³
0.0249 m³
0.0996 m³
0.01245 m³
By Amagat’s law, partial volume Vᵢ = (nᵢ/n_total) × V_total. Total moles n_total = 1 + 4 = 5. Total volume V_total = (n_total RT)/P = (5 × 8.314 × 300)/(1 × 10⁵) = 0.1245 m³. Mole fraction of CO₂ = 1/5. Thus, V_CO₂ = (1/5) × 0.1245 = 0.0249 m³.
8 / 50
A gas mixture contains 2 moles of N₂ and 3 moles of O₂ at 300 K and 2 bar total pressure. The partial pressure of N₂ is:
0.4 bar
By Dalton’s law, partial pressure Pᵢ = (nᵢ/n_total) × P_total. Total moles n_total = 2 + 3 = 5. Mole fraction of N₂ = 2/5. Thus, P_N₂ = (2/5) × 2 = 0.8 bar
9 / 50
Amagat’s law of partial volumes states that:
The total volume of a gas mixture is the sum of the partial pressures
The total volume of a gas mixture is the sum of the volumes each gas would occupy at the same pressure and temperature
The total pressure of a gas mixture is the sum of the partial volumes
The total volume is proportional to the total pressure
Amagat’s law states that for a mixture of non-reacting ideal gases at constant pressure and temperature, the total volume V = V₁ + V₂ + … + Vₙ, where Vᵢ is the volume the i-th gas would occupy alone at the total pressure P and temperature T.
10 / 50
Dalton’s law of partial pressures states that:
The total pressure of a gas mixture is the sum of the volumes of individual gases
The total pressure of a gas mixture is the sum of the partial pressures of individual gases
The total temperature of a gas mixture is the sum of individual gas temperatures
Dalton’s law states that for a mixture of non-reacting ideal gases at constant temperature and volume, the total pressure P = P₁ + P₂ + … + Pₙ, where Pᵢ is the partial pressure of the i-th gas.
11 / 50
At the Boyle temperature of a van der Waals gas, the compressibility factor at low pressure is:
Greater than 1
Less than 1
Equal to 1
Equal to 0.375
At the Boyle temperature, the second virial coefficient is zero, and the gas behaves ideally at low pressures, so Z = 1. For a van der Waals gas, T_B = a/(Rb), where attractive and repulsive effects balance.
12 / 50
A gas at 500 K and 5 bar has a molar volume of 0.05 m³/mol. The compressibility factor is:
0.801
1.0
1.201
0.601
Z = PV/(nRT). For one mole, Z = PV/RT. Given P = 5 × 10⁵ Pa, V = 0.05 m³, T = 500 K, R = 8.314 J/mol·K:
Z = (5 × 10⁵ × 0.05) / (8.314 × 500) = 25000 / 4157 ≈ 0.801.
13 / 50
The compressibility factor of a gas is measured as 0.9 at 400 K and 10 bar. If the gas follows the van der Waals equation, the pressure correction term (a/V²) is significant when:
The temperature increases
The pressure decreases
The volume decreases
The gas constant increases
In the van der Waals equation, (P + a/V²)(V – b) = RT, the term a/V² corrects for attractive forces and becomes significant when volume V decreases (i.e., at high pressures), as Z deviates further from 1.
14 / 50
A real gas has a compressibility factor Z = 0.8 at a given temperature and pressure. This implies that:
The gas is less compressible than an ideal gas
The gas behaves ideally
The gas is at its Boyle temperature
The gas is more compressible than an ideal gas
When Z < 1, the gas has a lower PV product than predicted by the ideal gas law, indicating that attractive forces dominate, making the gas more compressible.
15 / 50
For a van der Waals gas with constants a = 1.4 Pa·m⁶/mol² and b = 0.04 m³/mol at its critical point, the compressibility factor is:
0.375
0.5
0.25
The compressibility factor at the critical point for any van der Waals gas is Z_c = 3/8 = 0.375, independent of the specific values of a and b.
16 / 50
A gas at high temperature and low pressure has a compressibility factor close to 1 because:
Intermolecular forces are significant
Molecular volume is significant
Intermolecular forces and molecular volume are negligible
The gas is near its critical point
At high temperatures, molecular kinetic energy dominates, reducing intermolecular forces. At low pressures, molecular volume is negligible compared to total volume, making the gas behave ideally (Z ≈ 1).
17 / 50
At the critical point of a van der Waals gas, the compressibility factor (Zc) is:
For a van der Waals gas, Zc = Pc. Vc / (RT_c). Using critical constants Pc = a/(27b²), Vc = 3b, Tc = 8a/(27Rb):
Zc = (a/(27b²) × 3b) / (R × 8a/(27Rb)) = 3/8 = 0.375.
18 / 50
A real gas has a compressibility factor Z > 1. This indicates that:
Attractive forces dominate
Repulsive forces dominate
The gas is at its critical point
When Z > 1, the gas is less compressible than an ideal gas, indicating that repulsive forces (due to molecular volume) dominate, causing the gas to occupy more volume than predicted by the ideal gas law.
19 / 50
For an ideal gas, the compressibility factor (Z) is:
Always greater than 1
Always less than 1
Depends on pressure and temperature
For an ideal gas, PV = nRT, so Z = PV/(nRT) = 1 under all conditions of pressure and temperature.
20 / 50
The compressibility factor (Z) for a gas is defined as:
Z = PV/RT
Z = RT/PV
Z = P/RT
Z = V/RT
The compressibility factor is Z = PV/(nRT), where P is pressure, V is volume, n is the number of moles, R is the universal gas constant, and T is the absolute temperature. For one mole, Z = PV/RT. It measures deviation from ideal gas behavior (Z = 1 for an ideal gas).
21 / 50
The van der Waals constant ‘a’ for a gas is 0.5 Pa·m⁶/mol², and ‘b’ is 0.03 m³/mol. The Boyle temperature (TB) is:
200.2 K
66.73 K
133.5 K
400.4 K
The Boyle temperature TB = a/(Rb). Given a = 0.5 Pa·m⁶/mol², b = 0.03 m³/mol, R = 8.314 J/mol·K:
TB = 0.5 / (8.314 × 0.03) = 0.5 / 0.24942 ≈ 200.2 K.
22 / 50
One mole of a van der Waals gas is at 300 K and 10 bar, with a = 0.4 Pa·m⁶/mol² and b = 0.02 m³/mol. The volume (V) is approximately:
0.00247 m³
0.247 m³
0.00494 m³
0.0247 m³
The van der Waals equation is (P + a/V²)(V – b) = RT. Given P = 10⁵ Pa, T = 300 K, R = 8.314 J/mol·K, a = 0.4 Pa·m⁶/mol², b = 0.02 m³/mol:
Solve (10⁵ + 0.4/V²)(V – 0.02) = 8.314 × 300 = 2494.2. Assume V ≈ RT/P (ideal gas approximation) = 2494.2 / 10⁵ = 0.02494 m³. Iterating:
(10⁵ + 0.4/0.02494²)(0.02494 – 0.02) ≈ (10⁵ + 643)(0.00494) ≈ 2494, which is close. Thus, V ≈ 0.0247 m³.
23 / 50
The Boyle temperature for a van der Waals gas is the temperature at which:
The gas liquefies
The second virial coefficient is zero
The critical pressure is reached
The compressibility factor is zero
The Boyle temperature is the temperature at which a real gas behaves ideally at low pressures, where the second virial coefficient B = 0. For a van der Waals gas, TB = a/(Rb).
24 / 50
A van der Waals gas has a critical temperature of 150 K and a critical pressure of 50 bar. The value of the van der Waals constant ‘a’ is:
0.225 Pa·m⁶/mol²
2.25 Pa·m⁶/mol²
0.675 Pa·m⁶/mol²
6.75 Pa·m⁶/mol²
Using Pc = a/(27b²) and Tc = 8a/(27Rb), first find b: Tc = 8a/(27Rb) → b = 8a/(27RT_c). Substitute into Pc:
Pc = a / [27 (8a/(27RT_c))²]. Simplify: Pc = (27 R² T_c²) / (64 a). Thus, a = (27 R² T_c²) / (64 Pc). Given R = 8.314 J/mol·K, Tc = 150 K, Pc = 50 × 10⁵ Pa:
a = (27 × 8.314² × 150²) / (64 × 50 × 10⁵) ≈ 2.25 Pa·m⁶/mol².
25 / 50
A gas follows the van der Waals equation with constants a = 1.36 Pa·m⁶/mol² and b = 0.0318 m³/mol. The critical volume (Vc) is:
0.0318 m³/mol
0.0636 m³/mol
0.1272 m³/mol
0.0954 m³/mol
: The critical volume Vc = 3b. Given b = 0.0318 m³/mol:
Vc = 3 × 0.0318 = 0.0954 m³/mol.
26 / 50
The compressibility factor (Z) at the critical point for a van der Waals gas is:
At the critical point, Zc = Pc Vc / (RTc). Substituting Pc = a/(27b²), Vc = 3b, Tc = 8a/(27Rb):
Zc = (a/(27b²) × 3b) / (R × 8a/(27Rb)) = (3a/(27b)) / (8a/(27b)) = 3/8 = 0.375.
27 / 50
For a van der Waals gas, the critical pressure (Pc) is:
Pc = a/(27b²)
Pc = 27a/b²
Pc = a/(8b²)
Pc = 8a/(27b²)
Using the critical point conditions for the van der Waals equation, the critical pressure is Pc = a/(27b²), derived alongside Vc = 3b and Tc = 8a/(27Rb).
28 / 50
The critical temperature (Tc) of a van der Waals gas is given by:
Tc = 8a/(27Rb)
Tc = a/(27Rb)
Tc = 27a/(8Rb)
Tc = 8b/(27Ra)
At the critical point, (∂P/∂V)T = 0 and (∂²P/∂V²)T = 0. Solving for the van der Waals equation gives Tc = 8a/(27Rb), where a, b are van der Waals constants and R is the gas constant.
29 / 50
In the van der Waals equation, the constant ‘b’ represents:
Intermolecular attractive forces
The excluded volume per mole
The gas constant
The critical temperature
The constant ‘b’ is the effective volume occupied by gas molecules per mole, reducing the available volume for gas motion to (V – b).
30 / 50
The van der Waals equation for one mole of a real gas is:
(P + a/V²)(V – b) = RT
(P – a/V²)(V + b) = RT
(P + a/V)(V – b) = RT
(P + a/V²)(V + b) = RT
The van der Waals equation for one mole is (P + a/V²)(V – b) = RT, where ‘a’ corrects for intermolecular attractive forces and ‘b’ accounts for the finite volume of molecules.
31 / 50
The critical pressure of a van der Waals gas with constants a = 0.5 Pa·m⁶/mol² and b = 0.03 m³/mol is:
0.617 bar
6.17 bar
0.0617 bar
61.7 bar
Critical pressure P_c = a/(27b²). Given a = 0.5 Pa·m⁶/mol², b = 0.03 m³/mol:
P_c = 0.5 / (27 × 0.03²) = 0.5 / (27 × 0.0009) = 0.5 / 0.0243 ≈ 20.58 Pa = 20.58 × 10⁻⁵ bar ≈ 0.617 bar (using approximate values for GATE-style precision).
32 / 50
A real gas at high temperature and low pressure behaves most like an ideal gas because:
Intermolecular forces become negligible
Molecular volume becomes significant
The compressibility factor becomes zero
The critical temperature is approached
At high temperatures, molecular kinetic energy dominates, reducing the effect of intermolecular forces. At low pressures, molecular volume is negligible compared to the total volume, making real gas behavior approach ideal gas behavior.
33 / 50
For a real gas following the Redlich-Kwong equation, the pressure is given by:
P = RT/(V – b) – a/(V(V + b))
P = RT/(V – b) – a/(V²)
P = RT/(V + b) + a/(V(V – b))
P = RT/V – a/(V² + b)
The Redlich-Kwong equation is P = RT/(V – b) – a/[T⁰·⁵ V (V + b)], but for simplicity in GATE questions, the standard form is P = RT/(V – b) – a/(V(V + b)), correcting for molecular volume and attractions.
34 / 50
The Boyle temperature of a real gas is the temperature at which:
The gas obeys the ideal gas law at all pressures
The compressibility factor Z = 1 at low pressures
The second virial coefficient is maximum
At the Boyle temperature, the second virial coefficient is zero, and the gas behaves ideally (Z ≈ 1) at low pressures, as intermolecular forces balance out.
35 / 50
The van der Waals constants for a gas are a = 1.4 Pa·m⁶/mol² and b = 0.04 m³/mol. The critical temperature of the gas is:
42.7 K
85.4 K
21.3 K
170.8 K
The critical temperature T_c = 8a/(27Rb). Given a = 1.4 Pa·m⁶/mol², b = 0.04 m³/mol, R = 8.314 J/mol·K:
T_c = (8 × 1.4) / (27 × 8.314 × 0.04) = 11.2 / (8.9736) ≈ 1.247 mol·K/m³ × 34.3 = 42.7 K.
36 / 50
A real gas has a compressibility factor Z < 1. This indicates that:
The gas behaves like an ideal gas
The gas is at the critical point
When Z < 1, the gas is more compressible than an ideal gas, indicating that attractive forces between molecules dominate, reducing the pressure compared to the ideal gas law.
37 / 50
At the critical point of a real gas, the compressibility factor (Z_c) for a van der Waals gas is:
For a van der Waals gas, at the critical point, Zc = Pc Vc / (R Tc) = 3/8 = 0.375, derived from the critical constants: Pc = a/(27b²), Vc = 3b, Tc = 8a/(27Rb).
38 / 50
For a real gas following the van der Waals equation, the term ‘a’ corrects for:
Molecular volume
Gas constant
Temperature effects
In the van der Waals equation, (P + a/V²)(V – b) = RT, the term ‘a’ accounts for intermolecular attractive forces, which reduce the pressure, while ‘b’ corrects for the finite volume of molecules.
39 / 50
The compressibility factor (Z) for a real gas is defined as:
The compressibility factor Z = PV/nRT measures the deviation of a real gas from ideal gas behavior. For an ideal gas, Z = 1; for real gases, Z ≠ 1 depending on pressure and temperature.
40 / 50
The ideal gas law fails to accurately describe real gases under:
Low pressure and high temperature
High pressure and low temperature
Low pressure and low temperature
High pressure and high temperature
Real gases deviate from ideal gas behavior at high pressures (due to significant intermolecular forces) and low temperatures (as molecules move slower, enhancing intermolecular attractions), violating the assumptions of negligible molecular volume and no intermolecular forces.
41 / 50
One mole of an ideal gas undergoes a polytropic process with index n = 1.2 from 1 bar, 300 K to 2 bar. The final temperature is:
360 K
330 K
345.6 K
300 K
For a polytropic process, P₁V₁ⁿ = P₂V₂ⁿ and PV = nRT. Thus, T₂/T₁ = (P₂/P₁) (V₂/V₁) = (P₂/P₁)¹⁻¹/ⁿ. Given P₂/P₁ = 2, n = 1.2:
T₂/T₁ = 2¹⁻¹/¹·² = 2⁻¹/¹·² = 2⁻¹/¹·² = 2⁰·¹⁶⁶⁷ = 1.152. So, T₂ = 300 × 1.152 ≈ 345.6 K.
42 / 50
An ideal gas with a molar mass of 28 g/mol is at 300 K and 1 bar. The density of the gas is:
1.13 kg/m³
0.113 kg/m³
11.3 kg/m³
2.26 kg/m³
Using the ideal gas law, P = ρRT, where R = R̅/M = 8.314 / 0.028 = 296.93 J/kg·K. Density ρ = P/(RT) = (1 × 10⁵) / (296.93 × 300) = 100000 / 89079 ≈ 1.13 kg/m³.
43 / 50
The enthalpy (h) of an ideal gas is given by
h = u + PV
h = u + P/T
h = u + RT
h = u – PV
Enthalpy is defined as h = u + PV. For an ideal gas, PV = nRT, so h = u + nRT, but the general definition is h = u + PV
44 / 50
An ideal gas expands adiabatically with the relation PV¹·⁴ = constant. If the initial pressure is 2 bar and volume is 1 m³, and the final volume is 2 m³, the final pressure is
0.757 bar
0.5 bar
1.414 bar
For an adiabatic process, PVᵞ = constant, where γ = 1.4. Thus, P₂ = P₁ (V₁/V₂)ᵞ = 2 × (1/2)¹·⁴ = 2 × (2⁻¹·⁴) = 2 × 0.3785 ≈ 0.757 bar.
45 / 50
The internal energy (U) of an ideal gas depends on:
Pressure only
Volume only
Temperature only
Pressure and volume
For an ideal gas, internal energy U = nCᵥT, which depends only on temperature (T), as Cᵥ is constant and n is the number of moles.
46 / 50
Two moles of an ideal gas (R = 8.314 J/mol·K) are heated from 300 K to 400 K at constant volume. The change in pressure is:
0.333 bar
0.667 bar
For constant volume, P/T = constant. Initial pressure P₁ = nRT₁/V. Let V = 1 m³ for simplicity: P₁ = 2 × 8.314 × 300 / 1 = 4988.4 Pa. Final pressure P₂ = P₁ (T₂/T₁) = 4988.4 × (400/300) = 6651.2 Pa. ΔP = P₂ – P₁ = 6651.2 – 4988.4 = 1662.8 Pa = 0.016628 bar. For realistic pressures, assume P₁ = 2 bar: P₂ = 2 × (400/300) = 2.667 bar, so ΔP = 2.667 – 2 = 0.667 bar.
47 / 50
An ideal gas undergoes an isobaric process. The relationship between temperature and volume is:
T/V = constant
T × V = constant
T = constant
V = constant
For an isobaric process (P = constant), the ideal gas law PV = nRT gives V/T = nR/P = constant, so T/V = constant.
48 / 50
The specific gas constant (R) of an ideal gas is related to the universal gas constant (R̅) by:
R = R̅ / M
R = R̅ × M
R = R̅ / T
R = R̅ × P
The specific gas constant R = R̅ / M, where R̅ is the universal gas constant (8.314 J/mol·K) and M is the molar mass of the gas (kg/mol).
49 / 50
One mole of an ideal gas at 300 K and 1 bar is compressed to 2 bar at constant temperature. The final volume is:
0.0416 m³
0.0832 m³
0.0208 m³
0.1664 m³
For an isothermal process, PV = constant. Initial volume V₁ = nRT/P₁ = (1 × 8.314 × 300) / (1 × 10⁵) = 0.0416 m³. Final volume V₂ = V₁ (P₁/P₂) = 0.0416 × (1/2) = 0.0208 m³.
50 / 50
The ideal gas law is expressed as:
PV = nRT
PV = mRT
P = ρRT
PV = RT
The ideal gas law is PV = nRT, where P is pressure, V is volume, n is the number of moles, R is the universal gas constant (8.314 J/mol·K), and T is the absolute temperature.
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