Study of Gas Laws - Questions & Answers
1. PREVIOUS QUESTIONS
Short Answer Questions
1. Define or state – Boyle's Law
Answer: Boyle's Law states that at a constant temperature, the volume of a given mass of dry gas is inversely proportional to its pressure.
2. Express – Kelvin Zero in °C.
Answer: -273°C
Long Answer Questions
1. A fixed volume – of a gas occupies 760cm³ at 27°C & 70cm of Hg. What will be its vol. at s.t.p.
Answer:
Initial conditions: P1 = 70 cm Hg, V1 = 760 cm³, T1 = 27°C + 273 = 300 K.
Final conditions (S.T.P.): P2 = 76 cm Hg, T2 = 273 K.
Using the Gas Equation: (P1 × V1) / T1 = (P2 × V2) / T2
(70 × 760) / 300 = (76 × V2) / 273
V2 = (70 × 760 × 273) / (76 × 300) = 637 cm³.
2. At 0°C & 760 mm Hg pressure, a gas occupies a volume of 100 cm³. The Kelvin temperature [absolute temperature] of the gas is increased by one-fifth while the pressure is increased one & a half times. Calculate the final volume of the gas.
Answer:
Initial conditions: P1 = 760 mm Hg, V1 = 100 cm³, T1 = 0°C + 273 = 273 K.
Final conditions: P2 = 1.5 × 760 = 1140 mm Hg, T2 = 273 + (1/5 × 273) = 327.6 K.
Using the Gas Equation: (P1 × V1) / T1 = (P2 × V2) / T2
(760 × 100) / 273 = (1140 × V2) / 327.6
V2 = (760 × 100 × 327.6) / (1140 × 273) = 80 cc.
3. The pressure on one mole – of gas at s.t.p. is doubled and the temperature is raised to 546 K. What is the final volume of the gas? [one mole of a gas occupies a volume of 22.4 litres at stp.]
Answer:
Initial conditions: P1 = 1 atm, V1 = 22.4 litres, T1 = 273 K.
Final conditions: P2 = 2 atm, T2 = 546 K.
Using the Gas Equation: (P1 × V1) / T1 = (P2 × V2) / T2
(1 × 22.4) / 273 = (2 × V2) / 546
V2 = (1 × 22.4 × 546) / (2 × 273) = 22.4 litres.
4. Give a reason – "When stating the volume of a gas, the pressure & temperature should also be given"
Answer: The volume of a gas undergoes significant change if its pressure or temperature is slightly changed. Thus, stating the volume without standardizing or mentioning the pressure and temperature provides incomplete and meaningless information.
5. Is it possible to change the temperature and pressure of a fixed mass of gas without changing its volume. Explain your answer.
Answer: Yes, it is possible. According to the Gas Equation (P1×V1)/T1 = (P2×V2)/T2, if the volume remains constant (V1 = V2), the equation becomes P1/T1 = P2/T2. This means that if the absolute temperature and pressure are changed in the exact same proportion (e.g., both are doubled), the volume of the fixed mass of gas will remain unchanged.
2. ADDITIONAL QUESTIONS
1. What volume will a gas occupy at 740 mm pressure which at 1480 mm occupies 500 cc. [Temperature being constant]
Answer:
Using Boyle's Law: P1 × V1 = P2 × V2
1480 × 500 = 740 × V2
V2 = (1480 × 500) / 740 = 1000 cc.
2. The volume of a given mass of a gas at 27°C is 100 cc. To what temperature should it be heated at the same pressure so that it will occupy a volume of 150 cc.
Answer:
Using Charles' Law: V1 / T1 = V2 / T2
T1 = 27°C + 273 = 300 K.
100 / 300 = 150 / T2
T2 = (150 × 300) / 100 = 450 K.
In Celsius: 450 - 273 = 177°C.
3. A fixed mass of a gas has a volume of 750 cc at -23°C and 800 mm pressure. Calculate the pressure for which its volume will be 720 cc. The temperature being -3°C.
Answer:
Initial conditions: P1 = 800 mm, V1 = 750 cc, T1 = -23 + 273 = 250 K.
Final conditions: V2 = 720 cc, T2 = -3 + 273 = 270 K.
Using Gas Equation: (P1 × V1) / T1 = (P2 × V2) / T2
(800 × 750) / 250 = (P2 × 720) / 270
P2 = (800 × 750 × 270) / (720 × 250) = 900 mm.
4. What temperature would be necessary to double the volume of a gas initially at s.t.p. if the pressure is decreased by 50%.
Answer:
Initial conditions: P1 = P, V1 = V, T1 = 273 K.
Final conditions: P2 = 0.5P, V2 = 2V.
Using Gas Equation: (P1 × V1) / T1 = (P2 × V2) / T2
(P × V) / 273 = (0.5P × 2V) / T2
T2 = (0.5P × 2V × 273) / (P × V) = 1 × 273 = 273 K (or 0°C).
5. A gas cylinder having a capacity of 20 litres contains a gas at 100 atmos. How many flasks of 200 cm³ capacity can be filled from it at 1 atmos. pressure if the temperature remains constant.
Answer:
Using Boyle's Law: P1 × V1 = P2 × V2
100 × 20 = 1 × V2
V2 = 2000 litres.
Total volume available = 2000 litres = 2,000,000 cm³.
Number of flasks = 2,000,000 / 200 = 10,000 flasks.
6. A certain mass of gas occupied 850ml at a pressure of 760 mm of Hg. On increasing the pressure it was found that the volume of the gas was 75% of its initial volume. Assuming constant temperature, find the final pressure of the gas.
Answer:
Initial conditions: P1 = 760 mm Hg, V1 = 850 ml.
Final condition: V2 = 75% of 850 ml = 637.5 ml.
Using Boyle's Law: P1 × V1 = P2 × V2
760 × 850 = P2 × 637.5
P2 = (760 × 850) / 637.5 = 1013.33 mm of Hg.
7. It is required to reduce the volume of a gas by 20% by compressing it at a constant pressure. To do so, the gas is cooled. If the gas attains a final temperature of 157°C, find the initial temperature of the gas.
Answer:
Initial volume = V1. Final volume V2 = V1 - 0.20(V1) = 0.8 V1.
Final temperature T2 = 157 + 273 = 430 K.
Using Charles' Law: V1 / T1 = V2 / T2
V1 / T1 = 0.8 V1 / 430
T1 = 430 / 0.8 = 537.5 K.
In Celsius: 537.5 - 273 = 264.5°C.
8. At a given temperature the pressure of a gas reduces to 75% of its initial value & the volume increases by 40% of its initial value. Find this temperature if the initial temperature was -10°C.
Answer:
Initial conditions: P1 = P, V1 = V, T1 = -10 + 273 = 263 K.
Final conditions: P2 = 0.75P, V2 = 1.4V.
Using Gas Equation: (P1 × V1) / T1 = (P2 × V2) / T2
(P × V) / 263 = (0.75P × 1.4V) / T2
T2 = (0.75 × 1.4 × 263) / 1 = 1.05 × 263 = 276.15 K.
In Celsius: 276.15 - 273 = 3.15°C.
Multiple Choice Questions [MCQs] – Select the correct answer
1. State which of the following, represents the correct graphical relationship between – the volume of a given mass of dry gas to its absolute temperature at constant pressure.
(a) Graph A
(b) Graph B
(c) Graph C
(d) Graph D
Answer: (c) The straight line passing through the origin. According to Charles' Law, volume is directly proportional to absolute temperature (V ∝ T), which yields a straight line from the origin.
2. Match column A – with the relevant match in column B.
Column A
(a) 273 K
(b) V = K x 1/P
(c) 300 K
(d) V = K x T/P
Column B
(i) Gas equation
(ii) 27°C
(iii) 1 atmos. pressure
(iv) Boyles law
Answer:
(a) 273 K ➔ (iii) 1 atmos. pressure (representing combined S.T.P. conditions)
(b) V = K x 1/P ➔ (iv) Boyles law
(c) 300 K ➔ (ii) 27°C
(d) V = K x T/P ➔ (i) Gas equation
3. Assertion (A) : Gas laws are rules followed by a gas, on subjecting the gas to a change in – temperature, pressure or volume.
Reason (R) : Volume of gases are converted to s.t.p. conditions & then compared easily.
(a) Both A & R are true – and R is the correct explanation of A.
(b) Both A & R are true – but R is not the correct explanation of A.
(c) A is true – but R is false
(d) A is false – but R is true.
Answer: (b) Both A & R are true – but R is not the correct explanation of A.
High Order Thinking Question
1. Two gases 'A' & 'B' are compared at 273 K & 1 atmos. pressure to determine – which has a greater volume. The volume of gas 'A' is 1.20 litre at 27°C & 750 mm Hg & of gas 'B' 1.25 litre at 0°C & 760 mm. Hg. State which gas – 'A' or 'B' has a greater volume.
Answer:
Gas B is already at standard conditions (0°C/273K and 760 mm Hg/1 atm). Its volume is 1.25 litres.
For Gas A, converting to S.T.P.: P1 = 750 mm Hg, V1 = 1.20 litres, T1 = 27 + 273 = 300 K. P2 = 760 mm Hg, T2 = 273 K.
V2 = (P1 × V1 × T2) / (P2 × T1) = (750 × 1.20 × 273) / (760 × 300) = 1.077 litres.
Since 1.25 litres > 1.077 litres, Gas 'B' has a greater volume.
3. TEST – HIGH ORDER & CRITICAL THINKING QUESTIONS
UNIT TEST PAPER – 7
Q.1 Name or state the following :1. The law which states that pressure remaining constant the volume of a given mass of dry gas is directly proportional to its absolute [Kelvin] temperature.
Answer: Charles' Law
2. The law which studies the relationship between pressure of a gas and the volume occupied by it at constant temperature.
Answer: Boyle's Law
3. An equation used in chemical calculations which gives a simultaneous effect of changes of temperature and pressure on the volume of a given mass of dry gas.
Answer: Gas Equation
4. The standard pressure of a gas in cm. of mercury corresponding to one atmospheric pressure.
Answer: 76 cm of Hg
5. The absolute temperature value corresponding to 35°C.
Answer: 308 K
Q.2 Give reasons for the following :
1. Gases unlike solids and liquids exert pressure in all directions.
Answer: The molecules of a gas are in continuous random motion in straight lines with high velocities. They undergo random collisions with each other and strike the walls of the container, exerting an equal and uniform pressure in all directions.
2. Gases have lower densities compared to solids or liquids.
Answer: The intermolecular distance between gas molecules is very large. Consequently, the number of molecules per unit volume of a gas is much lower compared to solids and liquids, resulting in very low densities.
3. Temperature remaining constant the product of the vol. & the press. of a given mass of dry gas is a constant.
Answer: According to Boyle's law, volume is inversely proportional to pressure (V ∝ 1/P). Thus, P × V = constant. If the volume decreases, gas molecules have a shorter distance to strike the walls, increasing the collision frequency and thereby increasing the pressure proportionally to keep the product constant.
4. All temperatures on the Kelvin scale are in positive figures.
Answer: The Kelvin scale starts at absolute zero (-273°C), the theoretical lowest possible temperature where the volume of a gas is effectively zero. Because it is the absolute zero starting point, all other recorded temperatures are above it and are positive figures.
5. Volumes of gases are converted into s.t.p. conditions and then compared.
Answer: The volume of a gas undergoes major changes due to environmental fluctuations in temperature and pressure. Comparing gas volumes at standard, universal conditions (S.T.P.) provides a uniform baseline making comparisons mathematically accurate and easy.
Q.3 Calculate the following :
1. Calculate the temperature to which a gas must be heated, so that the volume triples without any change in pressure. The gas is originally at 57°C and having a volume 150 cc.
Answer:
Initial conditions: V1 = 150 cc, T1 = 57 + 273 = 330 K.
Final condition: V2 = 150 × 3 = 450 cc.
Using Charles' Law: V1 / T1 = V2 / T2
150 / 330 = 450 / T2
T2 = (450 × 330) / 150 = 990 K (or 717°C).
2. A gas 'X' at -33°C is heated to 127°C at constant pressure. Calculate the percentage increase in the volume of the gas.
Answer:
T1 = -33 + 273 = 240 K. T2 = 127 + 273 = 400 K.
Using Charles' Law: V1 / 240 = V2 / 400
V2 = (400 / 240) V1 = 1.667 V1.
Increase in volume = 1.667 V1 - V1 = 0.667 V1.
Percentage increase = 0.667 × 100 = 66.67%.
3. Calculate the volume of a gas 'A' at s.t.p., if at 37°C and 775mm of mercury pressure, it occupies a volume of 9½ litres.
Answer:
Initial conditions: P1 = 775 mm, V1 = 9.5 litres, T1 = 37 + 273 = 310 K.
Final conditions (S.T.P.): P2 = 760 mm, T2 = 273 K.
Using Gas Equation: V2 = (P1 × V1 × T2) / (P2 × T1) = (775 × 9.5 × 273) / (760 × 310) = 8.53 litres.
4. Calculate the temperature at which a gas 'A' at 20°C having a volume of 500 cc. will occupy a volume of 250 cc.
Answer:
Initial conditions: V1 = 500 cc, T1 = 20 + 273 = 293 K.
Final condition: V2 = 250 cc.
Using Charles' Law: V1 / T1 = V2 / T2
500 / 293 = 250 / T2
T2 = (250 × 293) / 500 = 146.5 K (or -126.5°C).
5. A gas 'X' is collected over water at 17°C and 750 mm. pressure. If the volume of the gas collected is 50 cc., calculate the volume of the dry gas at s.t.p. [at 17°C the vapour pressure is 14 mm.]
Answer:
Pressure of dry gas (P1) = Total Pressure - Vapour Pressure = 750 - 14 = 736 mm Hg.
Initial conditions: P1 = 736 mm, V1 = 50 cc, T1 = 17 + 273 = 290 K.
Final conditions (S.T.P.): P2 = 760 mm, T2 = 273 K.
Using Gas Equation: V2 = (P1 × V1 × T2) / (P2 × T1) = (736 × 50 × 273) / (760 × 290) = 45.58 cc.
Q.4 Assuming temperature remaining constant calculate the pressure of the gas in each of the following:
1. The pressure of a gas having volume 1000 cc. originally occupying 1500 cc. at 720 mm. pressure.
Answer: P1 = 720 mm, V1 = 1500 cc. V2 = 1000 cc. P2 = (720 × 1500) / 1000 = 1080 mm.
2. The pressure of a gas having volume 100 lits. originally occupying 75 dm³ at 700 mm. pressure.
Answer: Note: 75 dm³ = 75 litres. P1 = 700 mm, V1 = 75 litres. V2 = 100 litres. P2 = (700 × 75) / 100 = 525 mm.
3. The pressure of a gas having volume 380 lits. originally occupying 800 cm³ at 76 cm. pressure.
Answer: Note: 800 cm³ = 0.8 litres. P1 = 76 cm, V1 = 0.8 litres. V2 = 380 litres. P2 = (76 × 0.8) / 380 = 0.16 cm (or 1.6 mm).
4. The pressure of a gas having volume 1800 ml. originally occupying 300 ml. at 6 atms. pressure.
Answer: P1 = 6 atm, V1 = 300 ml. V2 = 1800 ml. P2 = (6 × 300) / 1800 = 1 atm.
5. The pressure of a gas having volume 1500 cm³ originally occupying 750 cc. at 5 atms. pressure.
Answer: Note: 1500 cm³ = 1500 cc. P1 = 5 atm, V1 = 750 cc. V2 = 1500 cc. P2 = (5 × 750) / 1500 = 2.5 atm.
Q.5 Calculate the following :
1. The temp. at which 500 cc. of a gas 'X' at temp. 293K occupies half it's original volume [pressure constant].
Answer: V1 = 500 cc, T1 = 293 K. V2 = 250 cc. T2 = (V2 × T1) / V1 = (250 × 293) / 500 = 146.5 K.
2. The volume at s.t.p. occupied by a gas 'Y' originally occupying 760 cc. at 300K and 70 cm. press. of Hg.
Answer: P1 = 70 cm, V1 = 760 cc, T1 = 300 K. P2 = 76 cm, T2 = 273 K. V2 = (70 × 760 × 273) / (76 × 300) = 637 cc.
3. The volume at s.t.p. occupied by a gas 'Z' originally occupying 1.57 dm³ at 310.5K and 75 cm. press. of Hg.
Answer: P1 = 75 cm, V1 = 1.57 dm³, T1 = 310.5 K. P2 = 76 cm, T2 = 273 K. V2 = (75 × 1.57 × 273) / (76 × 310.5) = 1.362 dm³.
4. The volume at s.t.p. occupied by a gas 'Q' originally occupying 153.7 cm³ at 287K and 750 mm. pressure [vapour pressure of gas 'Q' at 287K is 12 mm of Hg].
Answer: Dry gas pressure P1 = 750 - 12 = 738 mm Hg. V1 = 153.7 cm³, T1 = 287 K. P2 = 760 mm, T2 = 273 K. V2 = (738 × 153.7 × 273) / (760 × 287) = 141.96 cm³.
5. The temperature to which a gas 'P' has to be heated to triple it's volume, if the gas originally occupied 150 cm³ at 330K [pressure remaining constant].
Answer: V1 = 150 cm³, T1 = 330 K. V2 = 450 cm³. T2 = (V2 × T1) / V1 = (450 × 330) / 150 = 990 K.
Q.6 Fill in the blanks with the correct word, from the words in bracket:
1. If the temperature of a fixed mass of a gas is kept constant and the pressure is increased, the volume correspondingly decreases. [increases/decreases]
2. If the pressure of a fixed mass of a gas is kept constant and the temperature is increased, the volume correspondingly increases. [increases/decreases]
3. 1 dm³ of a gas is equal to 1 litre. [1 litre/100ml./100 cc.]
4. All the temperatures on the kelvin scale are in positive figures. [negative / positive]
5. At -273°C the volume of a gas is theoretically 0 cc. [272 cc./ 0 cc./ 274 cc.]