A gas has a pressure of 699.0 mm Hg at 40.0°C. What is the temperature at standard pressure?

1 Answer
Jul 6, 2017

Answer:

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"#. #larr# 100 kiloPascals

We need to convert #"mmHg"# to #"kPa"#.

#1"kPa"="7.5006 mmHg"#

#699.0color(red)cancel(color(black)("mmHg"))xx(1"kPa")/(7.5006color(red)cancel(color(black)("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_2T_1)/P_1#

#T_2=(100color(red)cancel(color(black)("kPa"))xx313"K")/(93.2color(red)cancel(color(black)("kPa")))="336 K"#

If you need to convert the temperature in Kelvins back to the Celsius temperature, subtract #273.15# from #"336 K"#.