Respuesta :
the answer:
When the volume of a gas is changed from 3.6 L to 15.5 L, the temperature will change from ?? oC to 87°C
application of charles law
charle's law tells that
T1/ V1 = T2 / V2, T must be in kelvin
it is given that V1 = 3.6 L, V2 = 15.5 L,
T2= 87°C= 360.15 K
so to find T1, T1 =(T2 / V2) x V1
T1= (360.15 / 15.5) * 3.6= 83.64° K = - 190.15° C
When the volume of a gas is changed from 3.6 L to 15.5 L, the temperature will change from ?? oC to 87°C
application of charles law
charle's law tells that
T1/ V1 = T2 / V2, T must be in kelvin
it is given that V1 = 3.6 L, V2 = 15.5 L,
T2= 87°C= 360.15 K
so to find T1, T1 =(T2 / V2) x V1
T1= (360.15 / 15.5) * 3.6= 83.64° K = - 190.15° C