Please solve these numericals
Answers
Please refer to the attachments.
1) initial
t1 = 25°c = 273+25 = 298k
p1 = 100cmHg = 1000mmHg
v1 = 80ml
final
V2 = 160ml
t2= 298 (since constant)
since temptature constant using Boyles law
p1v1 = p2v2
P2 = p1v1/V2 = 1000*80/160= 500mmhg = 50cmhg
2) initial
v1 = 74cm^3
p1= 760mmhg
Final
p2 = 740mmhg
Since temprature is constant so using Boyles law
p1v1 = p2 v2
v2 = p1v1/p2 = 74*760/740= 76cm^3
3) temptature is constant do Boyles law is used
Now
initial
p1= 100mmhg
v1 = 80cm^3
final
p2 = 125mmhg
v2 = X
using Boyles law
p1v1 = p2v2
v2 = p1v1/p2 = 100*80/125= 64cm^3
so
X= 64cm^3
again
initial
p1=200mmhg
v1=40cm^3
final
p2=y
V2=32cm^3
using Boyles law
p1v1 = p2v2
p2 = p1v1/v2
p2 = 200*40/32= 250mmhg or 250mm
4)initial
p1=760mmhg
v1=400cm^3
final
increase pressure = 25%= (1+25/100)760= 5*760/4= 950mmhg
p2=950mmhg
v2=?
using Boyles law
p1v1 = p2v2
v2 = p1v1/p2
v2 = 760*400/950 = 320cm^3
5)initial
p1= 330cmhg
v1= 600cm^3
final
since vessel A is connected to vessel B whose capacity is 300cm^3
so
v2 = 900cm^3
p2= ?
experiment is in vessel so temprature is constant
using Boyles law
p1v1 = p2v2
p2 = p1v1/v2 = 330*600/900= 220mmhg
6)initial
let initial volume be 100
v1 = 100cm^3
p1= 1080mmhg
final
since volume Decrease by 40%
so
v2 = 100*(1- 40/100)= 60cm^3
p2=?
since temprature is constant so using Boyles law
p1v1 = p2v2
p2 = p1v1/v2
p2 = 100*1080/60=1800mmhg