A sample of gas expands by 22.4L against an average pressure of 2.5atm. How much work (KJ) is done in the process? Is the work done by the system or by the surrounding?
Answers
When a gas expands at constant pressure then for a small change in volume
'
d
v
'
workdone is
d
w
=
p
d
v
.
If the volume changes from
'
v
1
'
to
'
v
2
'
at constant pressure
'
p
'
, the work done is
d
w
=
p
(
v
2
−
v
1
)
.
When work is done by the system against external pressure then
d
w
=
p
d
v
⇒
w
=
∫
v
2
v
1
p
d
v
So, by calculating with given numericals we get,
⇒
d
w
=
1.29
a
t
m
(
17.2
L
−
12.5
L
)
=
1.29
a
t
m
×
(
4.7
L
)
=
6.063
a
t
m
.
L
.
To have the answer in terms of we use below units:
1
a
t
m
=
101325
P
a
But
1
P
a
=
1
J
/
m
3
(
because
J
=
P
a
.
m
3
)
.
Substitute '
1
P
a
' value to get
'
1
a
t
m
' value,
⇒
1
a
t
m
=
101325
J
/
m
3
.
Now as we know
1
L
itre=
10
−
3
m
3
.
Converting the workdone
'
d
w
'
value into Joules using the above required units we get,
d
w
=
6.063
×
101325
J
m
3
×
10
−
3
m
3
=
614.3334
J
.
∴
work done in terms of Joules is
614.33
J
oules.