Question

a) Consider 1.3 moles of an ideal gas at an initial temperature of 400 K and in a 1.2 m³ closed container. If the gas goes through an isochoric process to twice the initial temperature, what is the new pressure of the gas in Pa?

b) Consider 1.3 moles of an ideal gas at an initial temperature of 400 K and in a 1.2 m³closed container. If the gas goes through an isothermal process to 3.6 m³, what is the new pressure of the gas in Pa?

c) Consider 1.3 moles of an ideal gas at an initial temperature of 400 K and in a 1.2 m³ closed container. If the gas goes through an isobaric process to 3.6 m³, what is the new temperature of the gas in Kelvin?

Answer 1

a)
Given
ideal gas , number of moles n = 1.3
initial temperature is T1 = 400 k
volume of the container v1 = 1.2 m^3
final temperature T2 = 2*T1 = 2*400 k = 800 k

new pressure of the gas P = ?

first we should find the pressure by using the ideal gas equation

   PV = n*R*T

   P = n*R*T/V

   P1 = 1.3*8.314*400/(1.2)

   P1 = 3602.733 Pa

now Isochoric process is at constant volume so

   P1/T1 = P2/T2

   P2 = P1*T2/T1

   P2 = 3602.733*800/400 Pa
   P2 = 7205.47 Pa
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b) ideal gas ,
number of moles n = 1.3
Temperature is = 400 k
volume of the container v1 = 1.2 m^3
final volume is V2 = 3.6 m^3

isothermal process, is at constant temperature

PV = n*R*T

   P = n*R*T/V

   P1 = 1.3*8.314*400/(1.2)

   P1 = 3602.733 Pa
From the relation, PV = constant

   P1*V1 = P2*V2

   P2 = P1*V1/V2
   P2 = 3602.733*1.2/(3.6) Pa
   P2 = 1201 Pa
c)
ideal gas ,
number of moles n = 1.3
Temperature is T1 = 400 k
volume of the container v1 = 1.2 m^3
final volume is V2 = 3.6 m^3

isobaric process, is at constant Pressure

PV = n*R*T

   P = n*R*T/V

   P1 = 1.3*8.314*400/(1.2)

   P1 = 3602.733 Pa

so the pressure is constant and

T2 =?

   from relation T1/V1 = T2/V2
T2 = T1*V2/V1 = 400*3.6/1.2 k = 1200 k

so New temperature is T2 = 1200 k