Ideal Gas Equations
1 / 10
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.
2 / 10
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.
3 / 10
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
4 / 10
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
1.0 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.
5 / 10
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.
6 / 10
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.
7 / 10
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.
8 / 10
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).
9 / 10
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.
10 / 10
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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