HEC USAT-E (Pre-Engineering) Chemistry Gases — Set 3

HEC USAT-E (Pre-Engineering) Chemistry Gases — Set 3

1. Question 1

   Q1. A gas has a density of 1.2 g/L at 1 atm and 27°C. What is its molecular weight?

   • A) 29.5
   • B) 30.0
   • C) 28.8
   • D) 32.0

   Answer: 29.5

   Explanation: Use d = PM / RT to find M: M = dRT / P = 1.2 * 0.0821 * 300 / 1 = 29.5 g/mol.

2. Question 2

   Q2. The partial pressure of a gas in a mixture is 0.2 atm. If the total pressure is 1 atm, what is its mole fraction?

   • A) 0.2
   • B) 0.5
   • C) 0.8
   • D) 0.1

   Answer: 0.2

   Explanation: The mole fraction is equal to the partial pressure divided by the total pressure: X = P_i / P_total = 0.2 / 1 = 0.2.

3. Question 3

   Q3. The ratio of the rates of diffusion of two gases is 2 : 1. What is the ratio of their molecular weights?

   • A) 1 : 4
   • B) 4 : 1
   • C) 1 : 2
   • D) 2 : 1

   Answer: 1 : 4

   Explanation: Graham's law: r1/r2 = √(M2/M1), so (r1/r2)² = M2/M1 = 4/1, hence M1/M2 = 1/4.

4. Question 4

   Q4. A gas occupies 2 L at 2 atm. What is its volume at 1 atm if the temperature remains constant?

   • A) 4 L
   • B) 1 L
   • C) 2 L
   • D) 8 L

   Answer: 4 L

   Explanation: Apply Boyle's law: P1V1 = P2V2, so V2 = P1V1 / P2 = 2 * 2 / 1 = 4 L.

5. Question 5

   Q5. The kinetic molecular theory assumes that the gas molecules

   • A) have a significant volume
   • B) attract each other strongly
   • C) are in constant random motion
   • D) have a high density

   Answer: are in constant random motion

   Explanation: The kinetic molecular theory assumes that gas molecules are in constant random motion, with no intermolecular forces.

6. Question 6

   Q6. The total pressure of a mixture of gases is the sum of the partial pressures of each gas. This is known as:

   • A) Dalton's Law
   • B) Henry's Law
   • C) Raoult's Law
   • D) Boyle's Law

   Answer: Dalton's Law

   Explanation: Dalton's Law states that P_total = P1 + P2 + ..., where P1, P2, etc., are partial pressures.

7. Question 7

   Q7. The solubility of a gas in a liquid is directly proportional to its partial pressure. This is stated by:

   • A) Henry's Law
   • B) Raoult's Law
   • C) Dalton's Law
   • D) Graham's Law

   Answer: Henry's Law

   Explanation: Henry's Law states that S = kP, where S is solubility and P is partial pressure.

8. Question 8

   Q8. The volume of a gas is directly proportional to the temperature in Kelvin at constant pressure. This is known as:

   • A) Charles' Law
   • B) Boyle's Law
   • C) Avogadro's Law
   • D) Gay-Lussac's Law

   Answer: Charles' Law

   Explanation: Charles' Law states that V1 / T1 = V2 / T2 at constant P.

9. Question 9

   Q9. Equal volumes of gases at the same temperature and pressure contain an equal number of molecules. This is stated by:

   • A) Avogadro's Law
   • B) Dalton's Law
   • C) Graham's Law
   • D) Boyle's Law

   Answer: Avogadro's Law

   Explanation: Avogadro's Law states that V ∝ n at constant T and P, where n is the number of moles.

10. Question 10

    Q10. The pressure of a gas is directly proportional to the temperature in Kelvin at constant volume. This is known as:

    • A) Gay-Lussac's Law
    • B) Charles' Law
    • C) Boyle's Law
    • D) Avogadro's Law

    Answer: Gay-Lussac's Law

    Explanation: Gay-Lussac's Law states that P1 / T1 = P2 / T2 at constant V.

11. Question 11

    Q11. The ratio of the rates of diffusion of two gases is equal to the inverse ratio of the square roots of their molecular weights. If the molecular weights of two gases are 16 and 64, the ratio of their rates of diffusion is:

    • A) 1 / 2
    • B) 2 / 1
    • C) 1 / 4
    • D) 4 / 1

    Answer: 2 / 1

    Explanation: Rate ratio = √(64 / 16) = √4 = 2 / 1.

12. Question 12

    Q12. A gas occupies 2 liters at 1 atm pressure. At constant temperature, its volume at 2 atm will be:

    • A) 4 liters
    • B) 1 liter
    • C) 2 liters
    • D) 0.5 liters

    Answer: 1 liter

    Explanation: P1V1 = P2V2, so 1 * 2 = 2 * V2, giving V2 = 1 liter.

13. Question 13

    Q13. The partial pressure of a gas in a mixture can be calculated using the formula:

    • A) P = (n / N) * P_total
    • B) P = (N / n) * P_total
    • C) P = n * P_total
    • D) P = N * P_total

    Answer: P = (n / N) * P_total

    Explanation: Partial pressure P = (n / N) * P_total, where n is the number of moles of the gas and N is the total number of moles.

14. Question 14

    Q14. At 0°C and 1 atm, 1 mole of an ideal gas occupies a volume of:

    • A) 11.2 liters
    • B) 22.4 liters
    • C) 33.6 liters
    • D) 44.8 liters

    Answer: 22.4 liters

    Explanation: At STP, 1 mole of an ideal gas occupies 22.4 liters.

15. Question 15

    Q15. The kinetic molecular theory of gases assumes that the molecules of a gas are:

    • A) stationary
    • B) in constant random motion
    • C) attracted to each other
    • D) in a fixed position

    Answer: in constant random motion

    Explanation: The kinetic theory assumes that gas molecules are in constant random motion.

16. Question 16

    Q16. The root mean square speed of the molecules of a gas is given by the equation:

    • A) √(3RT / M)
    • B) √(2RT / M)
    • C) √(RT / M)
    • D) √(3R / MT)

    Answer: √(3RT / M)

    Explanation: The RMS speed = √(3RT / M), where R is the gas constant, T is temperature, and M is molecular weight.

17. Question 17

    Q17. The behavior of real gases deviates from ideal gas behavior due to:

    • A) intermolecular forces and molecular size
    • B) only intermolecular forces
    • C) only molecular size
    • D) neither intermolecular forces nor molecular size

    Answer: intermolecular forces and molecular size

    Explanation: Real gases deviate from ideal behavior due to both intermolecular forces and the finite size of molecules.

18. Question 18

    Q18. The van der Waals equation for real gases is given by:

    • A) (P + a / V²)(V - b) = nRT
    • B) (P - a / V²)(V + b) = nRT
    • C) (P + a / V)(V - b) = nRT
    • D) (P - a / V)(V + b) = nRT

    Answer: (P + a / V²)(V - b) = nRT

    Explanation: The van der Waals equation accounts for intermolecular forces (a) and molecular size (b).

19. Question 19

    Q19. The critical temperature of a gas is the temperature above which it cannot be liquefied, no matter how high the pressure. This is because:

    • A) the molecules are moving too fast
    • B) the intermolecular forces are too weak
    • C) the gas is ideal
    • D) the pressure is too low

    Answer: the molecules are moving too fast

    Explanation: Above the critical temperature, the kinetic energy of the molecules overcomes the intermolecular forces, preventing liquefaction.

20. Question 20

    Q20. The liquefaction of gases is possible below the:

    • A) critical temperature
    • B) Boyle temperature
    • C) inversion temperature
    • D) triple point

    Answer: critical temperature

    Explanation: Gases can be liquefied below their critical temperature by applying sufficient pressure.