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Question 1Basic Principles

For the reaction 2A + 3B → C, a reactor is fed 4 mol of A and 3 mol of B. Which reactant limits the extent of reaction, and why?

  1. A, because it is fed in the smaller stoichiometric multiple
  2. Neither, because the feed mixture is exactly stoichiometric
  3. A, because its stoichiometric coefficient is larger
  4. B, because the feed ratio of B to A is below the stoichiometric ratio Correct answer.

Explanation

The stoichiometric requirement is 3 mol of B per 2 mol of A, i.e., a B/A ratio of 1.5. The actual feed ratio is 3/4 = 0.75, which is less than 1.5, so B runs out first and A remains in excess.

Question 2Basic Principles

An open tank holds water (density 1000 kg/m³) to a depth of 10 m. Taking g = 9.81 m/s², what is the gauge pressure at the tank bottom?

  1. About 980 kPa
  2. About 49 kPa
  3. About 98 kPa Correct answer.
  4. About 9.8 kPa

Explanation

For a static fluid, gauge pressure increases with depth according to p = ρgh. Substituting gives 1000 × 9.81 × 10 ≈ 98,100 Pa, or about 98 kPa.

Question 3Basic Principles

Which dimensionless group represents the ratio of inertial forces to viscous forces in pipe flow?

  1. Prandtl number
  2. Nusselt number
  3. Reynolds number Correct answer.
  4. Froude number

Explanation

The Reynolds number, Re = ρvD/μ, compares inertial to viscous forces and sets whether flow is laminar or turbulent. The other groups compare different transport phenomena.

Question 4Basic Principles

An adiabatic process is best described as one in which:

  1. no work crosses the system boundary
  2. the temperature of the system stays constant
  3. no heat crosses the system boundary Correct answer.
  4. the entropy of the system stays constant

Explanation

Adiabatic means the boundary is thermally insulated, so Q = 0 in the first law. Temperature and entropy may still change because of work or reaction.

Question 5Basic Principles

For a control volume at steady state with no chemical reaction, the general mass balance reduces to:

  1. total mass in equals total mass out Correct answer.
  2. total mass out equals mass generated
  3. total mass in plus total mass out equals zero
  4. accumulation equals total mass in

Explanation

At steady state nothing inside the control volume changes with time, so accumulation is zero and the mass entering must equal the mass leaving. Mass is conserved, so no generation term exists.

Question 6Basic Principles

What are the SI units of thermal conductivity?

  1. kg/(m·s)
  2. J/(kg·K)
  3. W/(m·K) Correct answer.
  4. W/(m²·K)

Explanation

Fourier's law q" = −k dT/dx shows that k has units of heat flux divided by temperature gradient, i.e., W/(m·K). The other options are units of a heat transfer coefficient, specific heat, and viscosity.

Question 7Basic Principles

Water flows at 2 kg/s and is heated from 20 °C to 60 °C. Taking cp = 4.18 kJ/(kg·K), what heat duty is required?

  1. 84 kW
  2. 334 kW Correct answer.
  3. 669 kW
  4. 167 kW

Explanation

For a single phase with constant heat capacity, Q = ṁcpΔT = 2 × 4.18 × 40 ≈ 334 kW. The temperature change is 40 K, not 60 K.

Question 8Basic Principles

Water with density 1000 kg/m³ and viscosity 1.0 mPa·s flows at 1.5 m/s through a pipe of 50 mm inside diameter. What is the Reynolds number?

  1. 1,500
  2. 75,000 Correct answer.
  3. 7,500
  4. 375,000

Explanation

Re = ρvD/μ = (1000 × 1.5 × 0.050)/0.001 = 75,000, which is well into the turbulent regime. The diameter must be converted to metres before substitution.

Question 9Basic Principles

Water flows steadily through a horizontal pipe that gradually contracts. Neglecting friction, what happens in the smaller-diameter section?

  1. The velocity decreases and the pressure rises
  2. Both the velocity and the pressure increase
  3. The velocity increases and the pressure falls Correct answer.
  4. Both the velocity and the pressure stay unchanged

Explanation

Continuity requires velocity to rise as the area falls. Bernoulli's equation for horizontal, frictionless flow then requires the pressure to fall to conserve mechanical energy.

Question 10Basic Principles

A pump moves 0.01 m³/s of water against a total head of 20 m with an overall efficiency of 70%. What shaft power is required?

  1. 2.8 kW Correct answer.
  2. 1.4 kW
  3. 4.0 kW
  4. 9.8 kW

Explanation

Hydraulic power is ρgQH = 1000 × 9.81 × 0.01 × 20 ≈ 1.96 kW. Dividing by the efficiency of 0.70 gives about 2.8 kW of shaft power.

Question 11Basic Principles

In a countercurrent heat exchanger the hot stream enters at 150 °C and leaves at 90 °C, while the cold stream enters at 30 °C and leaves at 70 °C. What is the log-mean temperature difference?

  1. About 70.0 °C
  2. About 69.5 °C Correct answer.
  3. About 60.5 °C
  4. About 75.0 °C

Explanation

The terminal differences are 150 − 70 = 80 °C and 90 − 30 = 60 °C. The LMTD is (80 − 60)/ln(80/60) ≈ 69.5 °C, slightly below the arithmetic mean.

Question 12Basic Principles

A first-order liquid-phase reaction with rate constant 0.10 min⁻¹ is carried out in a batch reactor for 20 min. What conversion is achieved?

  1. 86.5% Correct answer.
  2. 95.0%
  3. 20.0%
  4. 63.2%

Explanation

For a constant-volume first-order batch reaction, X = 1 − exp(−kt). With kt = 0.10 × 20 = 2.0, X = 1 − e⁻² ≈ 0.865.

Question 13Basic Principles

An ideal liquid mixture contains 50 mol% A. At the given temperature the vapor pressures of pure A and pure B are 100 kPa and 50 kPa. What is the bubble-point pressure of this mixture?

  1. 100 kPa
  2. 75 kPa Correct answer.
  3. 50 kPa
  4. 150 kPa

Explanation

At the bubble point the liquid is just saturated, so Raoult's law gives P = x_A P_A^sat + x_B P_B^sat. Substituting gives 0.5 × 100 + 0.5 × 50 = 75 kPa.

Question 14Basic Principles

A column separates A from B with a relative volatility of 2.5, giving a distillate of 95 mol% A and a bottoms of 5 mol% A. Using the Fenske equation, what is the minimum number of theoretical stages?

  1. About 12.8
  2. About 3.2
  3. About 6.4 Correct answer.
  4. About 2.5

Explanation

The Fenske equation gives N_min = ln[(x_D/(1−x_D))((1−x_B)/x_B)]/ln α. Here both composition ratios equal 19, so N_min = ln(361)/ln(2.5) ≈ 6.4.

Question 15Basic Principles

Heat is transferred from a hot fluid (film coefficient 1000 W/m²·K) through a thin wall of negligible resistance to a cold fluid (film coefficient 250 W/m²·K). What is the overall heat transfer coefficient based on the same area?

  1. 200 W/(m²·K) Correct answer.
  2. 750 W/(m²·K)
  3. 500 W/(m²·K)
  4. 1250 W/(m²·K)

Explanation

Film resistances add in series: 1/U = 1/h_i + 1/h_o = 0.001 + 0.004 = 0.005 m²·K/W. Inverting gives U = 200 W/(m²·K), controlled mainly by the smaller coefficient.

Question 16Basic Principles

How does the binary diffusion coefficient of a dilute gas pair behave?

  1. It rises with temperature and falls with pressure Correct answer.
  2. It falls as temperature rises at constant pressure
  3. It rises with pressure at constant temperature
  4. It is unaffected by both temperature and pressure

Explanation

Kinetic theory gives D proportional to T^1.5/P for gases, because higher temperature raises molecular speeds while higher pressure shortens the mean free path. Hence D increases with temperature and decreases with pressure.

Question 17Basic Principles

A liquid feed containing 100 mol/s of A enters an adiabatic reactor where A converts completely to B with a heat of reaction of −80 kJ/mol A. The reacting stream has a total heat capacity rate of 50 kW/K. Neglecting other effects, what is the adiabatic temperature rise?

  1. 160 K Correct answer.
  2. 16 K
  3. 320 K
  4. 80 K

Explanation

The heat released is 100 mol/s × 80 kJ/mol = 8000 kW. Setting this equal to the sensible heat capacity rate times the temperature rise gives ΔT = 8000/50 = 160 K.

Question 18Basic Principles

An ideal gas with a heat capacity ratio of 1.4 is compressed isentropically from 100 kPa and 300 K to 1000 kPa. What is the approximate outlet temperature?

  1. About 579 K Correct answer.
  2. About 1100 K
  3. About 450 K
  4. 300 K

Explanation

For isentropic compression of an ideal gas, T₂ = T₁(P₂/P₁)^((γ−1)/γ). With γ = 1.4 the exponent is 0.2857, so T₂ = 300 × 10^0.2857 ≈ 579 K.

Question 19Basic Principles

A proportional-only controller is applied to a self-regulating process and leaves a steady-state offset. Which change removes that offset?

  1. Reducing the proportional gain
  2. Adding derivative action
  3. Adding integral action Correct answer.
  4. Raising the proportional gain to a very high fixed value

Explanation

Integral action keeps changing the controller output as long as any error persists, so the error is driven to zero at steady state. Proportional action alone requires a nonzero error to hold a nonzero output.

Question 20Basic Principles

A project requires an installed capital of $500,000 and returns net cash savings of $125,000 per year for 10 years with no salvage value. At an interest rate of 10% per year, what is the net present worth?

  1. About $1,268,000
  2. About −$125,000
  3. About $750,000
  4. About $268,000 Correct answer.

Explanation

The present worth factor for a 10-year annuity at 10% is about 6.145, so the savings are worth roughly $768,000 today. Subtracting the $500,000 capital gives a net present worth of about $268,000.

Question 21Material and Energy Balances

How many gram-moles are contained in 36.0 g of water? (Molar mass of water = 18.0 g/mol.)

  1. 18.0 mol
  2. 36.0 mol
  3. 2.00 mol Correct answer.
  4. 0.500 mol

Explanation

Moles equal mass divided by molar mass: 36.0 g / 18.0 g/mol = 2.00 mol. A gram-mole is numerically identical to a mole when the mass is expressed in grams.

Question 22Material and Energy Balances

A tank holds 500 kg of an aqueous solution that is 10 wt% salt. What mass of salt does the tank contain?

  1. 50 kg Correct answer.
  2. 10 kg
  3. 500 kg
  4. 100 kg

Explanation

Mass of solute equals the mass fraction multiplied by the total mass: 0.10 x 500 kg = 50 kg. Weight percent is always on a total-mass basis.

Question 23Material and Energy Balances

An ideal gas is held at 0 degC and 1 atm, where its molar volume is 22.4 L/mol. How many moles are contained in 112 L of this gas?

  1. 5.00 mol Correct answer.
  2. 2.50 mol
  3. 0.200 mol
  4. 22.4 mol

Explanation

At 0 degC and 1 atm one mole of ideal gas occupies about 22.4 L. Dividing the volume by the molar volume gives 112/22.4 = 5.00 mol.

Question 24Material and Energy Balances

A steady stream of 1000 kg/h enters a splitter and leaves as two product streams. One product stream is 350 kg/h. What is the flow of the other product stream?

  1. 700 kg/h
  2. 350 kg/h
  3. 500 kg/h
  4. 650 kg/h Correct answer.

Explanation

With no reaction and no accumulation, total mass in equals total mass out. Subtracting the known product from the feed gives 1000 - 350 = 650 kg/h.

Question 25Material and Energy Balances

Under which condition does the accumulation term of a material balance on a process unit equal zero?

  1. The unit is operated adiabatically
  2. The reaction has reached chemical equilibrium
  3. The process operates at steady state Correct answer.
  4. The outlet streams are withdrawn as liquids

Explanation

Accumulation is the rate of change of material held within the system, and by definition it is zero when the process is at steady state. Reaction extent, heat transfer and phase of the outlets do not by themselves eliminate accumulation.

Question 26Material and Energy Balances

A liquid has a specific gravity of 1.2. Taking the density of water as 1.00 g/cm3, what is the density of the liquid in g/cm3?

  1. 0.83 g/cm3
  2. 1.2 g/cm3 Correct answer.
  3. 12 g/cm3
  4. 1200 g/cm3

Explanation

Specific gravity is the ratio of the liquid density to the density of water at the reference condition (1.00 g/cm3). Multiplying gives 1.2 x 1.00 = 1.2 g/cm3, which is also 1200 kg/m3.

Question 27Material and Energy Balances

Ethylene is hydrated to ethanol by C2H4 + H2O -> C2H5OH. A reactor is fed 100 mol/h of ethylene and 60 mol/h of steam, and the ethylene conversion is 50%. What is the ethanol production rate?

  1. 30 mol/h
  2. 60 mol/h
  3. 50 mol/h Correct answer.
  4. 100 mol/h

Explanation

Ethylene reacted is 0.50 x 100 = 50 mol/h, which needs 50 mol/h of steam and is satisfied by the 60 mol/h fed, so water is in excess. The 1:1 stoichiometry then gives 50 mol/h of ethanol.

Question 28Material and Energy Balances

Methane is burned with 25% excess air (air is 21 mol% O2 and 79 mol% N2). How many moles of air are supplied per mole of methane?

  1. 2.5 mol
  2. 11.9 mol Correct answer.
  3. 9.52 mol
  4. 10.0 mol

Explanation

Complete combustion, CH4 + 2O2 -> CO2 + 2H2O, requires 2 mol of O2 per mol of methane; 25% excess raises this to 2.5 mol of O2. Dividing by the oxygen content of air gives 2.5/0.21 = 11.9 mol of air.

Question 29Material and Energy Balances

A gas mixture containing 2.0 mol of methane and 3.0 mol of nitrogen is held at a total pressure of 5.0 bar. What is the partial pressure of methane?

  1. 1.0 bar
  2. 5.0 bar
  3. 3.0 bar
  4. 2.0 bar Correct answer.

Explanation

Dalton's law gives p_i = y_i P. The methane mole fraction is 2.0/5.0 = 0.40, so its partial pressure is 0.40 x 5.0 = 2.0 bar.

Question 30Material and Energy Balances

What heat duty is required to warm 10 kg/s of liquid water from 20 degC to 80 degC? The heat capacity is 4.18 kJ/(kg K).

  1. 251 kW
  2. 41.8 kW
  3. 418 kW
  4. 2510 kW Correct answer.

Explanation

For a single liquid phase with essentially constant heat capacity, Q = m cp dT = 10 x 4.18 x 60 = 2508 kW, about 2.51 MW.

Question 31Material and Energy Balances

For the water-gas shift reaction CO(g) + H2O(g) -> CO2(g) + H2(g), the standard heats of formation are CO = -110.5, H2O(g) = -241.8, CO2 = -393.5 and H2 = 0 kJ/mol. What is the standard heat of reaction?

  1. -283.0 kJ/mol
  2. -131.5 kJ/mol
  3. -41.2 kJ/mol Correct answer.
  4. +41.2 kJ/mol

Explanation

The heat of reaction is the sum of formation enthalpies of products minus that of reactants: (-393.5 + 0) - (-110.5 - 241.8) = -41.2 kJ/mol. The negative sign means the reaction releases heat.

Question 32Material and Energy Balances

A fresh feed of 100 mol/h of component A enters a process whose reactor is surrounded by a recycle loop. The overall conversion of A is 90% and the per-pass conversion in the reactor is 95%. What is the recycle flow of A?

  1. 200 mol/h
  2. 100 mol/h Correct answer.
  3. 50 mol/h
  4. 10 mol/h

Explanation

A leaving the process is 0.10 x 100 = 10 mol/h, and since all of it comes from the reactor effluent, the reactor effluent contains 10 mol/h of A. At 95% per-pass conversion the reactor feed of A is 10/0.05 = 200 mol/h, so the recycle is 200 - 100 = 100 mol/h.

Question 33Material and Energy Balances

A fresh feed of 100 mol/h containing 1.0 mol% inert enters a recycle loop. Inert leaves the process only in a purge stream that is held at 5.0 mol% inert. What purge flow is required at steady state?

  1. 100 mol/h
  2. 1 mol/h
  3. 5 mol/h
  4. 20 mol/h Correct answer.

Explanation

At steady state all inert entering must leave in the purge. Inert in the fresh feed is 0.010 x 100 = 1 mol/h, and if the purge is 5.0 mol% inert its total flow must be 1/0.05 = 20 mol/h.

Question 34Material and Energy Balances

A dryer receives 1000 kg/h of wet solids containing 20 wt% water and discharges solids containing 5 wt% water. How much water is removed?

  1. 50 kg/h
  2. 158 kg/h Correct answer.
  3. 150 kg/h
  4. 200 kg/h

Explanation

The bone-dry solids are 0.80 x 1000 = 800 kg/h and are unchanged by drying. The final wet mass is therefore 800/0.95 = 842 kg/h, so the water removed is 1000 - 842 = 158 kg/h.

Question 35Material and Energy Balances

A condenser must remove 500 kW. Cooling water enters at 20 degC and leaves at 35 degC, with a heat capacity of 4.18 kJ/(kg K). What cooling-water flow is needed?

  1. 3.99 kg/s
  2. 11.9 kg/s
  3. 0.125 kg/s
  4. 7.98 kg/s Correct answer.

Explanation

Solving Q = m cp dT for the flow gives m = 500/(4.18 x 15) = 7.98 kg/s. The 15 K rise in the cooling water is what sets the circulation rate.

Question 36Material and Energy Balances

Methane is burned completely with 20% excess air (21 mol% O2, 79 mol% N2). What is the mole fraction of oxygen in the dry flue gas?

  1. 16.7%
  2. 4.0%
  3. 2.9%
  4. 3.8% Correct answer.

Explanation

Per mole of methane, the 2.4 mol of O2 supplied leave 0.4 mol unreacted while 1 mol of CO2 forms, and the nitrogen carried in is 2.4 x 79/21 = 9.03 mol. The dry flue gas is 1 + 0.4 + 9.03 = 10.43 mol, so the O2 fraction is 0.4/10.43 = 0.038.

Question 37Material and Energy Balances

The standard heat of reaction at 25 degC is -50.0 kJ/mol. Over the temperature range of interest the difference in heat capacity between products and reactants is constant at -0.050 kJ/(mol K). What is the heat of reaction at 125 degC?

  1. -60.0 kJ/mol
  2. -50.0 kJ/mol
  3. -45.0 kJ/mol
  4. -55.0 kJ/mol Correct answer.

Explanation

Kirchhoff's law gives dH(T2) = dH(T1) + integral of dCp dT. With a constant dCp of -0.050 kJ/(mol K) over 100 K the correction is -5.0 kJ/mol, so the heat of reaction becomes -55.0 kJ/mol.

Question 38Material and Energy Balances

Steam at 100 degC (heat of vaporization 2257 kJ/kg) is sparged into 100 kg of water initially at 20 degC. The water heat capacity is 4.18 kJ/(kg K). What mass of steam condenses by the time the water reaches 60 degC?

  1. 6.9 kg Correct answer.
  2. 5.0 kg
  3. 8.3 kg
  4. 7.4 kg

Explanation

Heating the 100 kg of water by 40 K requires 100 x 4.18 x 40 = 16 720 kJ. Each kilogram of steam gives 2257 kJ on condensing plus 4.18 x 40 = 167 kJ on cooling from 100 to 60 degC, i.e. 2424 kJ/kg, so the mass needed is 16 720/2424 = 6.9 kg.

Question 39Material and Energy Balances

A vapor stream of 100 mol/h that is 50 mol% A and 50 mol% B is partially condensed. The liquid produced is 90 mol% B, and 40% of the feed is condensed. What is the mole fraction of A in the remaining vapor?

  1. 0.900
  2. 0.500
  3. 0.767 Correct answer.
  4. 0.233

Explanation

The liquid is 0.40 x 100 = 40 mol/h and holds 0.90 x 40 = 36 mol/h of B. The vapor therefore carries 50 - 36 = 14 mol/h of B in 60 mol/h, giving a B fraction of 0.233 and an A fraction of 1 - 0.233 = 0.767.

Question 40Material and Energy Balances

A fresh feed of 1000 mol/h contains 2.0 mol% inert with the balance being reactant A. Inert leaves the process only in a purge stream that is 10 mol% inert. How much A is lost in the purge?

  1. 20 mol/h
  2. 900 mol/h
  3. 180 mol/h Correct answer.
  4. 100 mol/h

Explanation

Inert entering is 0.020 x 1000 = 20 mol/h, and all of it exits in the purge, so the purge flow is 20/0.10 = 200 mol/h. The remaining 90% of the purge is A, giving 0.90 x 200 = 180 mol/h of A lost.

Question 41Thermodynamics

Using the convention that W is the work done by the system, the first law for a closed system with negligible changes in kinetic and potential energy reduces to which expression?

  1. Q + W = ΔU
  2. Q − W = ΔU Correct answer.
  3. Q = ΔH − W
  4. ΔU = W − Q

Explanation

For a closed system at rest, energy balance gives ΔU = Q − W when W is defined as work done by the system. Heat added raises internal energy, while work performed by the system removes energy. Enthalpy appears only when flow work is involved, as in open systems.

Question 42Thermodynamics

For a pure ideal gas, which of the following properties depends only on temperature?

  1. Density
  2. Entropy
  3. Internal energy Correct answer.
  4. Molar volume

Explanation

Because ideal-gas molecules have no intermolecular potential energy, internal energy (and enthalpy) is a function of temperature alone. Density, molar volume, and entropy all change with pressure as well as temperature for a fixed amount of gas.

Question 43Thermodynamics

A Carnot engine operates between a hot reservoir at 600 K and a cold reservoir at 300 K. What is its thermal efficiency?

  1. 0.50 Correct answer.
  2. 0.33
  3. 1.00
  4. 0.25

Explanation

Carnot efficiency depends only on the absolute reservoir temperatures: η = 1 − T_cold/T_hot. Substituting gives 1 − 300/600 = 0.50. No real engine operating between these reservoirs can exceed this value.

Question 44Thermodynamics

How many degrees of freedom exist for a single-component system in which two phases coexist at equilibrium?

  1. 2
  2. 1 Correct answer.
  3. 0
  4. 3

Explanation

The Gibbs phase rule is F = C − P + 2. With one component and two phases, F = 1 − 2 + 2 = 1, meaning that specifying one intensive variable (such as temperature) fixes the other (pressure).

Question 45Thermodynamics

For an isolated system, the total entropy change accompanying any spontaneous process is:

  1. Equal to the heat transferred
  2. Equal to zero
  3. Less than zero
  4. Greater than zero Correct answer.

Explanation

The second law states that the entropy of an isolated system increases for any real, irreversible process and stays constant only for a reversible one. Since a spontaneous process is irreversible, its total entropy change must be positive.

Question 46Thermodynamics

Which expression correctly represents the Clausius-Clapeyron relation for the temperature dependence of vapor pressure, assuming constant heat of vaporization?

  1. ln(P2/P1) = (ΔHvap/R)(1/T2 − 1/T1)
  2. ln(P2/P1) = (ΔHvap/R)(T2 − T1)
  3. ln(P2/P1) = −(ΔHvap/R)(1/T2 − 1/T1) Correct answer.
  4. P2/P1 = ΔHvap/(R T)

Explanation

Integrating d lnP/dT = ΔHvap/(RT²) between two temperatures gives ln(P2/P1) = −(ΔHvap/R)(1/T2 − 1/T1). The negative sign is required so that vapor pressure rises as temperature rises. The result can equivalently be written with (1/T1 − 1/T2).

Question 47Thermodynamics

An ideal gas is throttled adiabatically from 10 bar to 2 bar through a well-insulated valve. What happens to its temperature?

  1. It decreases
  2. It remains unchanged Correct answer.
  3. It increases
  4. It depends on the heat capacity ratio

Explanation

Throttling is isenthalpic because no work is done and the process is adiabatic. For an ideal gas, enthalpy is a function of temperature only, so constant enthalpy means constant temperature. Real gases may cool or warm depending on whether they lie below or above their Joule-Thomson inversion temperature.

Question 48Thermodynamics

For an exothermic gas-phase reaction at equilibrium, how does the equilibrium constant change when the temperature is raised at constant pressure?

  1. It first increases, then decreases
  2. It remains constant
  3. It increases
  4. It decreases Correct answer.

Explanation

The van't Hoff relation d lnK/dT = ΔH°/(RT²) shows that K rises with temperature only when ΔH° is positive. For an exothermic reaction ΔH° is negative, so lnK and therefore K decrease as temperature increases. Le Chatelier's principle gives the same conclusion: heating shifts an exothermic equilibrium toward reactants.

Question 49Thermodynamics

For a pure ideal gas, the fugacity coefficient at any pressure is:

  1. Unity Correct answer.
  2. Numerically equal to the pressure in bar
  3. The reciprocal of the pressure
  4. Zero

Explanation

The fugacity coefficient is defined by φ = f/P, and for an ideal gas the fugacity equals the pressure, so φ = 1. Departures of φ from unity measure nonideality in a real gas. Because ideal-gas behavior is defined without intermolecular forces, φ is unity at every pressure for that model.

Question 50Thermodynamics

A gas at 300 K and 50 bar has a compressibility factor of 0.85. What is its molar volume?

  1. 5.87 × 10⁻⁴ m³/mol
  2. 4.24 × 10⁻⁴ m³/mol Correct answer.
  3. 3.60 × 10⁻⁴ m³/mol
  4. 4.99 × 10⁻⁴ m³/mol

Explanation

The compressibility factor definition PV = ZRT gives V = ZRT/P. Using P = 5.0 × 10⁶ Pa, V = (0.85)(8.314)(300)/(5.0 × 10⁶) = 4.24 × 10⁻⁴ m³/mol. The ideal-gas volume ZRT/P with Z = 1 would be larger, as expected for Z below unity.