30 pUmm Problems: 4 questions: Practice Questions Fill in the Blanks (I point ea
ID: 546181 • Letter: 3
Question
30 pUmm Problems: 4 questions: Practice Questions Fill in the Blanks (I point each): Kinetic Molecular Theory Assumptionts and continuous random motion. 1). Molecules are in The average kinetic energy of a substance is determined by the The molecule of a gas are 2). What is the effect on the volume of one mole of an ideal gas with no between them. a). The pressure is reduced by a factor of four at constant temperature. b). The pressure changes from 760 torr to 202 kPa and the temperature changes from 37 °C to 155K e). The pressure changes from 2 atm to 101 kPa and the temperature changes from 305 K to 32°C SHORT ANSWER 1). Use the following terminology (System and surroundings) to define an endothermic and an exothermic reaction? (2) Page 2Explanation / Answer
(1) Molecules are in constant, rapid and continous random motion.
The average kinetic energy of a substance is determined by the temperature.
The molecules of a gas are small size, no volume with no attraction between them.
(2) As we know the ideal gas equation,
PV=nRT .......................(1)
(a) Here Temperature is constant.
Given, P2 = P1/4 (As the pressure is reduced four times) ........................(2)
Here P1 = initial pressure
P2 = final pressure
So, we can write equation (1), as
P1V1 = P2V2 .....................(3)
Putting equation (2) in (3), we get
P1V1 = (P1/4)V2
=> V1 = V2/4
=> V2 = 4V1
Hence the volume of the ideal gas is increased by a factor of four.
(b) Here, all the factors are varying like pressure, temperature and volume. So, from the equation (1), we can write
P1V1/T1 = P2V2/T2 ...................(4)
Given, P1 = 760 torr =101.325 kPa ~ 101 kPa T1 = 37o C = 37+273.15 = 310.15 K ~ 310 K
P2 = 202 kPa T2 = 155 K
Putting these values in (4), we get
(101)V1/(310) = (202)V2/(155)
V2/V1 = 155x101/310x202 = 1/4
V2 = V1/4
Hence, volume of the ideal gas is decreased by a factor of four.
(c) Given,
P1 = 2 atm =202.650 kPa ~ 202 kPa T1 = 305 K
P2 = 101 kPa T2 = 32o C = 32+273.15 = 305.15 K ~ 305 K
Putting in equation (4), we get
(202)V1/(305) = (101)V2/(305)
V2/V1 = 305x202/305x101 = 2
V2 = 2V1
Hence the volume of the ideal gas is increased by a factor of two.
Short Answer:
An exothermic reaction is the one where enengy released from system to its surrounding.
An endothermic reaction is the one where enengy absorbed from surrounding to the system.