A gas has a pressure of 699 mm Hg at 40.0 °C. What is the temperature at standard pressure of 760 mmHg?
Answers
The temperature will be
336 K
.
Explanation:
This is a pressure-temperature gas problem. This means that it involves Gay-Lussac's law, which states that the pressure of a given amount of a gas, held at constant volume, varies directly with its temperature. This means that if the pressure increases, so does the temperature, and vice-versa. The equation used to solve this problem is:
P
1
T
1
=
P
2
T
2
Before we go further, we need to determine what standard pressure is, and we need to convert the Celsius temperature to Kelvin temperature by adding
273.15
to the Celsius temperature.
Standard pressure is
100 kPa
.
←
100 kiloPascals
We need to convert
mmHg
to
kPa
.
1
kPa
=
7.5006 mmHg
699.0
mmHg
×
1
kPa
7.5006
mmHg
=
93.2 kPa
Organize your data:
Known
P
1
=
93.2 kPa
T
1
=
40
∘
C
+
273.15
=
313 K
P
2
=
100 kPa
Unknown
T
2
=
?
Solution
Rearrange the equation above to isolate
T
2
. Insert the given data into the new equation and solve.
T
2
=
P
2
T
1
P
1
T
2
=
100
kPa
×
313
K
93.2
kPa
=
336 K
If you need to convert the temperature in Kelvins back to the Celsius temperature, subtract
273.15
from
336 K
.