The pressure of a gas at S.T.P is tripled when temperature is raised to 546 C ∘ . What will be the change in volume?

S.T.P. are the standard temperature and pressure condition. The standard temperature is 273 K ( 0 ∘ Celsius) and the standard pressure is 1 atm Pressure. At S.T.P, one mole of gas occupies 22 . 4 L Of volume (molar volume). Given : Pressure P 1 = 1 atm Volume V 1 = 22 . 4 L Temperature T 1 = 273 K Pressure P 2 = 3 ( P 1 ) = 3 × 1 = 3 atm Temperature T 2 ( K ) = 546 + 273 = 819 K To find: Volume V 2 = ? Formula used: P 1 V 1 T 1 = P 2 V 2 T 2 Calculation: 1 × 22 . 4 273 = 3 × V 2 819 V 2 = 1 × 22 . 4 × 819 3 × 273 V 2 = 22 . 4 L

The pressure of a gas at S.T.P is tripled when temperature is raised to 546 C ∘ . What will be the change in volume?

S.T.P. are the standard temperature and pressure condition. The standard temperature is 273 K ( 0 ∘ Celsius) and the standard pressure is 1 atm Pressure. At S.T.P, one mole of gas occupies 22 . 4 L Of volume (molar volume). Given : Pressure P 1 = 1 atm Volume V 1 = 22 . 4 L Temperature T 1 = 273 K Pressure P 2 = 3 ( P 1 ) = 3 × 1 = 3 atm Temperature T 2 ( K ) = 546 + 273 = 819 K To find: Volume V 2 = ? Formula used: P 1 V 1 T 1 = P 2 V 2 T 2 Calculation: 1 × 22 . 4 273 = 3 × V 2 819 V 2 = 1 × 22 . 4 × 819 3 × 273 V 2 = 22 . 4 L

Diffusion is the process of movement of molecules under a concentration gradient. It is the movement of molecules from a region of higher concentration to a region of lower concentration down the concentration gradient. It mostly happens in liquids and gases because their particles move randomly from place to place.

Boyle’s law states that the pressure exerted by a gas is inversely proportional to the volume occupied by it. In other words, the pressure and volume of a gas are inversely proportional to each other as long as the temperature and the quantity of gas are kept constant. Boyle’s law was put forward by the Anglo-Irish chemist Robert Boyle in the year 1662. It is expressed mathematically as: Pα 1 V PV = K Hence, Boyle's law equation is P 1 V 1 = P 2 V 2

Charles' law states that the volume of an ideal gas is directly proportional to the absolute temperature at constant pressure. This law was formulated in the year 1780 by French physicist Jacques Charles. It can be represented in the following manner. V ∝ T V T = K Hence, Charles' law equation is at two different volume and temperature can be expressed as: V 1 T 1 = V 2 T 2

Match the following: Column I (Temperature in Celsius) Column II (Temperature in Kelvin) (i) 0 C ∘ (a) 546 K (ii) 273 C ∘ (b) 293 K (iii) - 273 C ∘ (c) 290 K (iv) 20 C ∘ (d) 0 K (v) 17 C ∘ (e) 273 K

The Kelvin, unit of thermodynamic temperature, is the fraction 1 273 . 16 of the thermodynamic temperature of the triple point of water. The unit of Celsius temperature is the degree Celsius, symbol C ∘ , which is by definition equal in magnitude to the Kelvin. The relationship between Celsius and Kelvin is as follows, K = ° C + 273 Therefore, K = 0 + 273 K = 273 Temperature is measured in Celsius ° C , Fahrenheit ° F and Kelvin K . Kelvin is the SI unit of temperature. The relationship between Celsius and Kelvin is as follows, K = ° C + 273 K = 273 + 273 K = 546 The absolute zero is the lowest limit of temperature at which the volume becomes theoretically zero. The temperature for absolute zero is - 273 ° C . K = ° C + 273 K = - 273 + 273 K = 0 20 ° C is equal to 293 K , as the K = ° C + 273 . The standard temperature is 0 ° C or 273 K . According to relation between Celsius and Kelvin, K = ° C + 273 . Therefore 17 ° C is equal to 290 K . Column I (Temperature in Celsius) Column II (Temperature in Kelvin) 0 C ∘ 273 K 273 C ∘ 546 K - 273 C ∘ 0 K 20 C ∘ 293 K 17 C ∘ 290 K

E M B I B E

TOGETHER WITH CHEMISTRY>Chapter 7 - Study of Gas Laws>PRACTICE QUESTIONS>Q 9

EASY

9th ICSE

IMPORTANT

Earn 100

The pressure of a gas at S.T.P is tripled when temperature is raised to $546 {}^{\circ}C$ . What will be the change in volume?

Important Questions on Study of Gas Laws

MEDIUM

9th ICSE

IMPORTANT

TOGETHER WITH CHEMISTRY>Chapter 7 - Study of Gas Laws>PRACTICE QUESTIONS>Q 10

Why kelvin temperature is always positive?

EASY

9th ICSE

IMPORTANT

TOGETHER WITH CHEMISTRY>Chapter 7 - Study of Gas Laws>PRACTICE QUESTIONS>Q 11

Define diffusion.

MEDIUM

9th ICSE

IMPORTANT

TOGETHER WITH CHEMISTRY>Chapter 7 - Study of Gas Laws>PRACTICE QUESTIONS>Q 12

Explain why absolute zero is a theoretical concept?

MEDIUM

9th ICSE

IMPORTANT

TOGETHER WITH CHEMISTRY>Chapter 7 - Study of Gas Laws>Self Evaluation Test>Q 1

State Boyle's law.

EASY

9th ICSE

IMPORTANT

TOGETHER WITH CHEMISTRY>Chapter 7 - Study of Gas Laws>Self Evaluation Test>Q 1

Define Charles' law.

MEDIUM

9th ICSE

IMPORTANT

TOGETHER WITH CHEMISTRY>Chapter 7 - Study of Gas Laws>Self Evaluation Test>Q 2

Name the quantity which is constant in

(i) Boyle's law

(ii) Charles' law

MEDIUM

9th ICSE

IMPORTANT

TOGETHER WITH CHEMISTRY>Chapter 7 - Study of Gas Laws>Self Evaluation Test>Q 3

What will be the pressure required to reduce

600 mL

of a dry gas at

750 mm

pressure to

500 mL

at the same temperature.

MEDIUM

9th ICSE

IMPORTANT

TOGETHER WITH CHEMISTRY>Chapter 7 - Study of Gas Laws>Self Evaluation Test>Q 4

Match the following:

Column I (Temperature in Celsius)	Column II (Temperature in Kelvin)
(i) $0 {}^{\circ}C$	(a) $546 K$
(ii) $273 {}^{\circ}C$	(b) $293 K$
(iii) $- 273 {}^{\circ}C$	(c) $290 K$
(iv) $20 {}^{\circ}C$	(d) $0 K$
(v) $17 {}^{\circ}C$	(e) $273 K$