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KSEAB II PUC Chemistry study materials
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KSEAB II PUC Chemistry practice questions

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Question 1Some Basic Concepts of Chemistry

How many significant figures are present in the measured value 0.004560 g?

  1. 7
  2. 3
  3. 5
  4. 4 Correct answer.

Explanation

Leading zeros that only locate the decimal point are never significant, so the zeros in 0.004 are ignored. The digits 4, 5, 6 and the trailing zero after 6 are all significant, giving four significant figures.

Question 2Some Basic Concepts of Chemistry

One mole of any substance contains 6.022 x 10^23 entities. How many oxygen atoms are present in 0.50 mol of ozone, O3?

  1. 3.011 x 10^23
  2. 6.022 x 10^23
  3. 9.033 x 10^23 Correct answer.
  4. 1.807 x 10^24

Explanation

Each ozone molecule contains three oxygen atoms, so 0.50 mol of O3 contains 1.50 mol of oxygen atoms. Multiplying 1.50 mol by Avogadro's number gives about 9.03 x 10^23 atoms.

Question 3Some Basic Concepts of Chemistry

The molecular formula of a compound is twice its empirical formula. If the empirical formula mass is 30 u and the molar mass is 60 g/mol, what is the molecular formula when the empirical formula is CH2O?

  1. CH2O
  2. C6H12O6
  3. C3H6O3
  4. C2H4O2 Correct answer.

Explanation

The ratio n equals molar mass divided by empirical formula mass, that is 60/30 = 2. Multiplying each subscript of CH2O by 2 gives C2H4O2.

Question 4Some Basic Concepts of Chemistry

In the reaction N2(g) + 3H2(g) -> 2NH3(g), 28 g of nitrogen reacts with 6 g of hydrogen. Which statement correctly describes the limiting reagent and the mass of ammonia formed?

  1. N2 is limiting and 17 g of NH3 forms
  2. H2 is limiting and 34 g of NH3 forms
  3. Both are limiting and 68 g of NH3 forms
  4. N2 is limiting and 34 g of NH3 forms Correct answer.

Explanation

28 g of N2 is 1 mol and 6 g of H2 is 3 mol, which is exactly the 1:3 stoichiometric ratio, so neither reagent is in excess and both are consumed. One mole of N2 gives 2 mol of NH3, that is 34 g, and the same answer follows from hydrogen, confirming N2 as the reference limiting reagent here.

Question 5Some Basic Concepts of Chemistry

Which of the following represents the correct number of significant figures for the result of 12.11 + 18.0 + 1.013, reported to the proper precision?

  1. 31
  2. 31.12
  3. 31.1 Correct answer.
  4. 31.123

Explanation

For addition and subtraction, the answer is rounded to the same number of decimal places as the measurement with the fewest decimal places. Here 18.0 has one decimal place, so the sum 31.123 is rounded to 31.1.

Question 6Some Basic Concepts of Chemistry

A gaseous hydrocarbon contains 85.7% carbon and 14.3% hydrogen by mass. What is its empirical formula?

  1. C2H4
  2. C2H6
  3. CH2 Correct answer.
  4. CH4

Explanation

Dividing each percentage by atomic mass gives 85.7/12 = 7.14 mol of C and 14.3/1 = 14.3 mol of H. The ratio H:C is 14.3/7.14 = 2, so the simplest whole-number formula is CH2.

Question 7Some Basic Concepts of Chemistry

What volume of 0.20 M HCl is needed to neutralise exactly 25.0 mL of 0.10 M Na2CO3 according to Na2CO3 + 2HCl -> 2NaCl + H2O + CO2?

  1. 6.25 mL
  2. 50.0 mL
  3. 25.0 mL Correct answer.
  4. 12.5 mL

Explanation

Moles of Na2CO3 equal 0.0250 L x 0.10 M = 2.50 x 10^-3 mol. The equation shows 2 mol HCl per mol carbonate, so 5.00 x 10^-3 mol HCl is required, and dividing by 0.20 M gives 0.0250 L, that is 25.0 mL.

Question 8Some Basic Concepts of Chemistry

A sample of an oxide of iron contains 69.9% iron and 30.1% oxygen by mass. What is its empirical formula?

  1. FeO
  2. FeO2
  3. Fe3O4
  4. Fe2O3 Correct answer.

Explanation

Moles of Fe = 69.9/55.85 = 1.25 and moles of O = 30.1/16.0 = 1.88. Dividing both by 1.25 gives Fe:O = 1:1.5, and multiplying by 2 gives the simplest whole-number ratio Fe2O3.

Question 9Structure of Atom

The number of orbitals present in the third shell (n = 3) of an atom is

  1. 4
  2. 16
  3. 9 Correct answer.
  4. 3

Explanation

For a shell with principal quantum number n, the number of orbitals equals n squared. Substituting n = 3 gives 9 orbitals, distributed as one 3s, three 3p and five 3d orbitals. The maximum electron capacity of this shell is therefore 18.

Question 10Structure of Atom

Which of the following sets of quantum numbers is permissible for an electron in an atom?

  1. n = 1, l = 1, m = 0, s = +1/2
  2. n = 2, l = 2, m = 0, s = +1/2
  3. n = 4, l = 1, m = -2, s = +1/2
  4. n = 3, l = 2, m = -2, s = -1/2 Correct answer.

Explanation

The azimuthal quantum number l may take values from 0 up to n - 1, and m must lie between -l and +l. For n = 3 the allowed l values are 0, 1, 2, and with l = 2 the value m = -2 is within range, so this set is valid.

Question 11Structure of Atom

The energy of an electron in a hydrogen atom depends solely on its

  1. spin quantum number s
  2. azimuthal quantum number l
  3. principal quantum number n Correct answer.
  4. magnetic quantum number m

Explanation

In the Bohr model of the hydrogen atom the energy is given by E = -13.6/n squared eV, so it depends only on the principal quantum number. Because hydrogen has a single electron, there is no shielding to split subshells of the same n, so l, m and s do not affect the energy in this model.

Question 12Structure of Atom

The de Broglie wavelength of an electron accelerated through a potential difference V is proportional to

  1. V
  2. 1 divided by the square root of V Correct answer.
  3. V squared
  4. the square root of V

Explanation

Kinetic energy gained is eV, so momentum p = square root of (2meV). Since the de Broglie wavelength equals h/p, it is proportional to 1 divided by the square root of V. Increasing the accelerating voltage therefore shortens the wavelength.

Question 13Structure of Atom

Which transition in the hydrogen atom emits a photon of the highest energy?

  1. n = 3 to n = 2
  2. n = 4 to n = 3
  3. n = 5 to n = 4
  4. n = 2 to n = 1 Correct answer.

Explanation

The energy gap between successive levels decreases as n increases, since E is proportional to -1/n squared. The largest gap among the choices is between n = 2 and n = 1, so this transition releases the most energetic photon.

Question 14Structure of Atom

According to Hund's rule, the correct ground-state configuration of the nitrogen atom (Z = 7) has

  1. three 2p electrons with two paired and one with opposite spin
  2. two 2p electrons paired and one unpaired
  3. three 2p electrons singly occupying all three 2p orbitals with parallel spins Correct answer.
  4. three 2p electrons paired in one orbital

Explanation

Hund's rule states that degenerate orbitals are filled singly with parallel spins before any pairing occurs. Nitrogen has three 2p electrons, so each of the three 2p orbitals receives one electron with the same spin, giving a half-filled subshell.

Question 15Structure of Atom

The ground-state electron configuration of chromium (Z = 24) is best written as

  1. [Ar] 3d4 4s2
  2. [Ar] 3d3 4s3
  3. [Ar] 3d5 4s1 Correct answer.
  4. [Ar] 3d6 4s0

Explanation

A half-filled 3d subshell with a single 4s electron gives extra exchange stability, so chromium adopts [Ar] 3d5 4s1 rather than the simple aufbau prediction. This is the experimentally observed ground-state configuration.

Question 16Structure of Atom

The number of unpaired electrons in the ground state of the Fe3+ ion (Z of Fe = 26) is

  1. 3
  2. 5 Correct answer.
  3. 4
  4. 2

Explanation

Neutral iron is [Ar] 3d6 4s2. Removing three electrons gives Fe3+ as [Ar] 3d5, and by Hund's rule the five 3d electrons occupy all five orbitals singly. This half-filled d5 arrangement has five unpaired electrons.

Question 17Structure of Atom

In the quantum mechanical model, the square of the wavefunction, psi squared, at a point represents

  1. the probability density of finding the electron there Correct answer.
  2. the exact path followed by the electron
  3. the energy of the electron at that point
  4. the velocity of the electron at that point

Explanation

Born's interpretation states that psi squared gives the probability density of locating the electron at a given point. This replaced the idea of definite orbits with the concept of an orbital, a region of high probability.

Question 18Structure of Atom

The energy difference between successive Bohr orbits of the hydrogen atom decreases as n increases. This directly explains why

  1. the first ionization energy of hydrogen is zero
  2. the electron can occupy only the n = 1 level
  3. the spectral lines in a series converge to a limit at high n Correct answer.
  4. all hydrogen lines have identical wavelengths

Explanation

As n grows, adjacent levels lie closer in energy, so transitions ending on a fixed lower level produce lines whose spacing shrinks. The lines therefore crowd together and converge to a series limit.

Question 19Structure of Atom

An electron and a proton are accelerated through the same potential difference. The ratio of their de Broglie wavelengths (electron to proton) is closest to

  1. about 43 Correct answer.
  2. about 0.023
  3. about 1836
  4. 1

Explanation

Both particles gain the same kinetic energy eV, so wavelength is proportional to 1 divided by the square root of mass. The proton is about 1836 times heavier than the electron, so the ratio of wavelengths equals the square root of 1836, roughly 43.

Question 20Structure of Atom

The magnetic quantum number m describes

  1. the orientation of an orbital in space Correct answer.
  2. the size of an orbital
  3. the spin of an electron
  4. the shape of an orbital

Explanation

The magnetic quantum number takes values from -l to +l and specifies how an orbital of given l is oriented in space. Its name comes from the splitting of spectral lines in an applied magnetic field.

Question 21Chemical Bonding and Molecular Structure

According to VSEPR theory, the shape of the NH₄⁺ ion is best described as:

  1. Square planar
  2. See-saw
  3. Trigonal pyramidal
  4. Tetrahedral Correct answer.

Explanation

The ammonium ion has four bonding pairs around nitrogen and no lone pairs; the four electron pairs repel to positions of maximum separation, giving a bond angle near 109.5° and a regular tetrahedral geometry.

Question 22Chemical Bonding and Molecular Structure

The hybridisation of the carbon atoms in the ethyne (C₂H₂) molecule is:

  1. sp Correct answer.
  2. sp²
  3. sp³
  4. sp³d

Explanation

Each carbon in ethyne forms two sigma bonds (one to hydrogen, one to the other carbon) and two pi bonds, so only two hybrid orbitals are needed; mixing one s and one p orbital gives sp hybridisation with linear geometry.

Question 23Chemical Bonding and Molecular Structure

Which of the following molecules is non-polar despite containing polar bonds?

  1. NH₃
  2. CO₂ Correct answer.
  3. HCl
  4. H₂O

Explanation

In CO₂ the two C=O bond dipoles are equal and point in exactly opposite directions along the linear axis, so they cancel and the net dipole moment is zero. Symmetry, not bond polarity alone, decides the molecular dipole.

Question 24Chemical Bonding and Molecular Structure

The number of sigma and pi bonds respectively in the HCN molecule is:

  1. 2 and 2 Correct answer.
  2. 2 and 1
  3. 3 and 1
  4. 1 and 2

Explanation

Hydrogen cyanide contains one C–H single bond and one C≡N triple bond. A single bond contributes one sigma bond, and a triple bond contributes one sigma plus two pi bonds, giving two sigma and two pi bonds in total.

Question 25Chemical Bonding and Molecular Structure

Which species has a bond order of 2.5 according to molecular orbital theory?

  1. O₂⁻
  2. N₂
  3. O₂⁺ Correct answer.
  4. O₂

Explanation

For O₂⁺ the electron configuration gives eight bonding and three antibonding electrons, so bond order = (8 − 3)/2 = 2.5. Removing an electron from an antibonding orbital of O₂ increases the bond order above 2.

Question 26Chemical Bonding and Molecular Structure

The bond angle in H₂O (104.5°) is smaller than that in NH₃ (107°) mainly because:

  1. The hydrogen atoms in water are closer together due to hydrogen bonding between molecules
  2. Oxygen is more electronegative than nitrogen, so the O–H bonds are shorter
  3. Water has two lone pairs on oxygen while ammonia has one, giving stronger lone pair–bond pair repulsion Correct answer.
  4. Ammonia is planar while water is bent

Explanation

Lone pairs occupy more space than bonding pairs, so each additional lone pair on the central atom compresses the bond angle. Water has two lone pairs compared with one in ammonia, so its H–O–H angle is reduced more.

Question 27Chemical Bonding and Molecular Structure

Which of the following molecules has a linear shape?

  1. NO₂⁻
  2. SO₂
  3. SnCl₂
  4. BeCl₂ Correct answer.

Explanation

Beryllium chloride has two bonding pairs and no lone pairs on beryllium, so the electron pairs arrange at 180° and the molecule is linear. Two-coordinate species with lone pairs instead bend.

Question 28Chemical Bonding and Molecular Structure

In the BF₃ molecule, the boron atom is sp² hybridised. The resulting molecular geometry and approximate bond angle are:

  1. Trigonal planar, 120° Correct answer.
  2. Linear, 180°
  3. Tetrahedral, 109.5°
  4. Trigonal pyramidal, 107°

Explanation

Three sp² hybrid orbitals on boron point towards the corners of an equilateral triangle, so BF₃ is trigonal planar with F–B–F angles of about 120°. There is no lone pair on boron to distort the arrangement.

Question 29Chemical Bonding and Molecular Structure

The dipole moment of NH₃ is greater than that of NF₃. The most likely reason is that:

  1. Nitrogen is more electronegative than fluorine
  2. In NF₃ the lone pair dipole opposes the bond dipoles, while in NH₃ it reinforces them Correct answer.
  3. N–F bonds are less polar than N–H bonds
  4. NF₃ is planar while NH₃ is pyramidal

Explanation

In ammonia the bond dipoles and the lone-pair dipole point in the same direction, so they add. In nitrogen trifluoride the strongly polar N–F dipoles point away from the lone pair, so they partly cancel it, giving a smaller net moment.

Question 30Chemical Bonding and Molecular Structure

Which of the following species is isoelectronic with the cyanide ion, CN⁻?

  1. NO⁺
  2. All three listed species Correct answer.
  3. N₂
  4. CO

Explanation

The cyanide ion has 14 electrons (6 from carbon, 7 from nitrogen, 1 from the charge). N₂, CO and NO⁺ each also contain 14 electrons, so all three are isoelectronic with CN⁻ and share a triple-bond description.

Question 31Chemical Bonding and Molecular Structure

Which of the following contains both ionic and covalent bonding?

  1. CH₄
  2. NaCl
  3. H₂O
  4. NH₄Cl Correct answer.

Explanation

Ammonium chloride is built from NH₄⁺ and Cl⁻ ions held by ionic attraction, while within the ammonium ion the nitrogen–hydrogen bonds are covalent. Such species are said to show both types of bonding.

Question 32Chemical Bonding and Molecular Structure

The bond order of the O₂⁻ ion (superoxide) and the number of unpaired electrons it contains are respectively:

  1. 2.5 and 1
  2. 1.5 and 2
  3. 1.5 and 1 Correct answer.
  4. 2.0 and 0

Explanation

Adding one electron to O₂ places it in a pi antibonding orbital, giving seven bonding and four antibonding electrons: bond order = (7 − 4)/2 = 1.5. The single electron in the pi antibonding set leaves one unpaired electron.

Question 33Chemical Bonding and Molecular Structure

The percentage s-character in the hybrid orbitals used by carbon for the C–H bonds in ethyne, ethene and ethane respectively is:

  1. 50, 25, 33
  2. 33, 50, 25
  3. 25, 33, 50
  4. 50, 33, 25 Correct answer.

Explanation

An sp hybrid orbital has one s and one p orbital, so its s-character is 1/2 = 50%; sp² gives 1/3 ≈ 33%; and sp³ gives 1/4 = 25%. Greater s-character shortens and strengthens the bond, which is why ethyne C–H bonds are the shortest.

Question 34Chemical Bonding and Molecular Structure

The formal charges on the central oxygen, the singly bonded oxygen and the doubly bonded oxygen in the ozone molecule (O₃) are respectively:

  1. −1, +1, 0
  2. 0, −1, +1
  3. +1, −1, 0 Correct answer.
  4. +1, 0, −1

Explanation

Formal charge equals valence electrons minus non-bonding electrons minus half the bonding electrons. The central oxygen has three bonds and one lone pair, giving 6 − 2 − 3 = +1; the singly bonded terminal oxygen has three lone pairs, giving 6 − 6 − 1 = −1; the doubly bonded oxygen gives 6 − 4 − 2 = 0.

Question 35Chemical Bonding and Molecular Structure

Which of the following statements about sigma and pi bonds is correct?

  1. A double bond consists of two sigma bonds
  2. Pi bonds are formed by sideways overlap of p orbitals and are weaker than sigma bonds Correct answer.
  3. Sigma bonds can be formed only by overlap of p orbitals
  4. Pi bonds allow free rotation about the bond axis

Explanation

A pi bond arises from lateral overlap of parallel p orbitals, giving a diffuse electron cloud above and below the internuclear axis; this overlap is less effective than the head-on overlap of a sigma bond, so pi bonds are weaker and restrict rotation.

Question 36Chemical Bonding and Molecular Structure

Which of the following correctly lists the species in order of increasing bond length?

  1. N₂ < O₂ < F₂ Correct answer.
  2. O₂ < N₂ < F₂
  3. N₂ < F₂ < O₂
  4. F₂ < O₂ < N₂

Explanation

Bond order decreases from N₂ (3) to O₂ (2) to F₂ (1), and bond length increases as bond order falls. Hence N₂ has the shortest bond and F₂ the longest, giving the order N₂ < O₂ < F₂.

Question 37States of Matter and Thermodynamics

A sealed rigid vessel contains an ideal gas at constant temperature. If the amount of gas is doubled while the volume is kept constant, what happens to the pressure?

  1. It is halved.
  2. It remains unchanged.
  3. It becomes four times the original value.
  4. It is doubled. Correct answer.

Explanation

For an ideal gas at fixed temperature and volume, pressure is directly proportional to the number of moles (Avogadro's law combined with the ideal gas equation pV = nRT). Doubling n at constant V and T therefore doubles p.

Question 38States of Matter and Thermodynamics

Which statement correctly describes the deviation of a real gas from ideal behaviour at high pressure?

  1. The compression factor Z becomes less than 1 because attractive forces dominate.
  2. The compression factor Z becomes greater than 1 because molecular volume is significant. Correct answer.
  3. The gas obeys pV = nRT exactly at all pressures.
  4. The compression factor Z equals 1 because repulsive forces vanish.

Explanation

At high pressure the molecules are forced close together, so the finite volume occupied by the molecules themselves becomes important. This makes the gas harder to compress than an ideal gas, giving Z greater than 1.

Question 39States of Matter and Thermodynamics

Equal masses of oxygen (molar mass 32 g/mol) and helium (molar mass 4 g/mol) are placed in identical containers at the same temperature. Which gas exerts the greater pressure?

  1. Helium, because it has more moles for the same mass. Correct answer.
  2. Both exert equal pressure because the masses are equal.
  3. Oxygen, because its molecules are heavier.
  4. Oxygen, because it has a larger molar mass.

Explanation

Pressure at fixed V and T depends on the number of moles, not mass. Equal mass of helium corresponds to eight times as many moles as oxygen, so helium exerts the greater pressure.

Question 40States of Matter and Thermodynamics

The critical temperature of a gas is the temperature above which the gas:

  1. liquefies on applying any pressure.
  2. has zero compressibility factor.
  3. behaves exactly as an ideal gas at all pressures.
  4. cannot be liquefied however great the pressure. Correct answer.

Explanation

Above the critical temperature the kinetic energy of the molecules is too high for attractive forces to hold them in a liquid state. No amount of pressure can liquefy the gas above this temperature.

Question 41States of Matter and Thermodynamics

According to Graham's law, the rate of diffusion of a gas is:

  1. inversely proportional to its molar mass.
  2. inversely proportional to the square root of its molar mass. Correct answer.
  3. directly proportional to the square root of its molar mass.
  4. directly proportional to its molar mass.

Explanation

Graham's law states that the rate of diffusion of a gas is inversely proportional to the square root of its molar mass at constant temperature and pressure. Lighter gases therefore diffuse faster.

Question 42States of Matter and Thermodynamics

A gas mixture contains 2 mol of nitrogen and 3 mol of oxygen at a total pressure of 5 atm. What is the partial pressure of oxygen?

  1. 5 atm
  2. 3 atm Correct answer.
  3. 1 atm
  4. 2 atm

Explanation

By Dalton's law the partial pressure of a component equals its mole fraction times the total pressure. The mole fraction of oxygen is 3/5, so its partial pressure is (3/5) × 5 atm = 3 atm.

Question 43States of Matter and Thermodynamics

Which of the following is a correct statement of the first law of thermodynamics for a closed system?

  1. ΔH = q always, for any process.
  2. ΔU = q + w, where w is work done on the system. Correct answer.
  3. ΔU = 0 for every spontaneous process.
  4. ΔU = q − w, where w is work done on the system.

Explanation

The first law expresses conservation of energy: the change in internal energy equals heat added to the system plus work done on the system. Using the sign convention where w is work on the system, ΔU = q + w.

Question 44States of Matter and Thermodynamics

For an isothermal reversible expansion of an ideal gas, which quantity is zero?

  1. The change in entropy of the gas.
  2. The heat absorbed by the gas.
  3. The work done by the gas.
  4. The change in internal energy. Correct answer.

Explanation

The internal energy of an ideal gas depends only on temperature. In an isothermal process the temperature is constant, so ΔU is zero and q equals the work done.

Question 45States of Matter and Thermodynamics

Which of the following processes has a negative value of ΔS for the system?

  1. Sublimation of iodine.
  2. Melting of ice at 0 °C.
  3. Condensation of steam to water. Correct answer.
  4. Dissolution of ammonium chloride in water.

Explanation

Entropy decreases when a gas condenses to a liquid because the molecules become more ordered and the number of accessible microstates falls. The other listed processes increase disorder.

Question 46States of Matter and Thermodynamics

For a reaction to be spontaneous at all temperatures, which combination of signs of ΔH and ΔS is required?

  1. ΔH positive and ΔS negative.
  2. ΔH positive and ΔS positive.
  3. ΔH negative and ΔS positive. Correct answer.
  4. ΔH negative and ΔS negative.

Explanation

Spontaneity requires ΔG = ΔH − TΔS to be negative. With ΔH negative and ΔS positive, both terms favour a negative ΔG at every temperature, so the reaction is spontaneous throughout.

Question 47States of Matter and Thermodynamics

The standard enthalpy of formation of an element in its most stable form at 298 K and 1 bar is:

  1. always negative.
  2. equal to its molar mass in kJ.
  3. equal to its standard entropy.
  4. zero by convention. Correct answer.

Explanation

The standard enthalpy of formation of an element in its reference state is defined as zero at 298 K and 1 bar. This convention allows enthalpies of formation of compounds to be tabulated and used in Hess's law calculations.

Question 48States of Matter and Thermodynamics

A system absorbs 500 J of heat and does 200 J of work on the surroundings. What is the change in internal energy of the system?

  1. 300 J Correct answer.
  2. 700 J
  3. −700 J
  4. −300 J

Explanation

Using ΔU = q + w with work done on the system positive, work done by the system is w = −200 J. Thus ΔU = 500 J + (−200 J) = 300 J.

Question 49States of Matter and Thermodynamics

Which relation correctly connects enthalpy change and internal energy change at constant pressure for a reaction involving gases, treating them as ideal?

  1. ΔU = ΔH + ΔnRT
  2. ΔH = ΔU − ΔnRT
  3. ΔH = ΔU + RT
  4. ΔH = ΔU + ΔnRT Correct answer.

Explanation

At constant pressure, ΔH = ΔU + Δ(pV). For ideal gases Δ(pV) = ΔnRT, where Δn is the change in moles of gas, giving ΔH = ΔU + ΔnRT.

Question 50States of Matter and Thermodynamics

Which of the following is true for a reversible process in a Carnot cycle operating between two temperatures?

  1. The efficiency can exceed 1 if the gas is ideal.
  2. The efficiency is 1 − T_cold/T_hot with temperatures in kelvin. Correct answer.
  3. The efficiency depends on the working substance used.
  4. The efficiency is independent of the temperature difference.

Explanation

The Carnot efficiency of a reversible engine is 1 − T_cold/T_hot when temperatures are expressed in kelvin. It is the maximum possible efficiency for any engine operating between those reservoirs and does not depend on the working substance.