Question

A sample of helium occupies 160cm³ at 100kPa and 25°C. What volume will it occupy if the pressure is adjusted to 80kPa and if the temperature remains unchanged?

Answer 1

PV=nRT

Where:

P
P is the pressure of the gas (in Pascals)
V
V is the volume of the gas (in cubic meters)
n
n is the number of moles of the gas
R
R is the ideal gas constant (
8.314
 
J/mol
⋅
K
8.314J/mol⋅K)
T
T is the temperature of the gas (in Kelvin)
Since we're given that the temperature remains unchanged, we can use the relationship
P
1
V
1
T
1
=
P
2
V
2
T
2
T
1
​

P
1
​
V
1
​

​
=
T
2
​

P
2
​
V
2
​

​
, where
P
1
V
1
P
1
​
V
1
​
and
P
2
V
2
P
2
​
V
2
​
represent the initial and final conditions, respectively, and
T
1
=
T
2
T
1
​
=T
2
​
because the temperature remains constant.

Given:

P
1
=
100
 
kPa
=
100000
 
Pa
P
1
​
=100kPa=100000Pa
V
1
=
160
 
cm
3
=
0.00016
 
m
3
V
1
​
=160cm
3
=0.00016m
3

P
2
=
80
 
kPa
=
80000
 
Pa
P
2
​
=80kPa=80000Pa
We need to find
V
2
V
2
​
.

P
1
V
1
T
1
=
P
2
V
2
T
2
T
1
​

P
1
​
V
1
​

​
=
T
2
​

P
2
​
V
2
​

​


Since
T
1
=
T
2
T
1
​
=T
2
​
, we can cancel them out:

P
1
V
1
=
P
2
V
2
P
1
​
V
1
​
=P
2
​
V
2
​


Now, solve for
V
2
V
2
​
:

V
2
=
P
1
V
1
P
2
V
2
​
=
P
2
​

P
1
​
V
1
​

​


V
2
=
100000
×
0.00016
80000
V
2
​
=
80000
100000×0.00016
​


V
2
=
16
8
V
2
​
=
8
16
​


V
2
=
0.00032
 
m
3
V
2
​
=0.00032m
3


So, the volume of helium will be
0.00032
 
m
3
0.00032m
3
when the pressure is adjusted to 80 kPa while keeping the temperature constant.