Answer:
293°C.
Explanation:
We're asked to find the new temperature of the
NH
3
after it is subdued to a change in volume (with constant pressure).
To do this, we can use the temperature-volume relationship of gases, illustrated by Charles's law:
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
∣
∣
∣
V
1
T
1
=
V
2
T
2
∣
∣
∣
−−−−−−−−−−−−−−
(constant pressure and quantity of gas)
where
V
1
and
V
2
are the initial and final volumes of the gas
T
1
and
T
2
are the initial and final absolute temperatures of the gas (which must be in Kelvin)
We have:
V
1
=
25
L
V
2
=
50
L
T
1
=
10
o
C
+
273
=
283
l
K
−−−−−
T
2
=
?
Let's rearrange the above equation to solve for the final temperature,
T
2
:
T
2
=
V
2
T
1
V
1
Plugging in known values:
T
2
=
(
50
L
)
(
283
l
K
)
25
L
=
566
l
K
−−−−−
You asked for the temperature in degrees Celsius, so we convert back:
566
l
K
−
273
=
¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯¯
∣
∣
293
l
o
C
∣
∣
−−−−−−−−−−−−
