Question

Calculate the temperature of a gas when `"6.00 L"` at `20.0^@"C"` is compressed to `"4.00L"`?

i have trouble figuring out which gas law i need to use. I know you have to turn C into Kelvin by adding 273.
I just need help with the equations!

Answer 1

The temperature of the compressed gas is `"195 K"`.

Explanation:

The variables are temperature and volume, so you will use the temperature-volume law , or Charles' law , which states that the volume of a given amount of gas held at constant pressure is directly proportional to its temperature. This means that if the volume increases, so does the temperature, and vice versa. The equation is:

`V_1/T_1=V_2/T_2`,

where:

`V_1` and `T_1` are the initial volume and temperature, and `V_2` and `T_2` are the final volume and temperature.

Known

`V_1="6.00 L"`

`T_1="20.0"^@"C + 273.15"="293.2 K"`

`V_2="4.00 L"`

Unknown

`T_2`

Solution

Rearrange the equation to isolate `T_2`.

`T_2=(V_2T_1)/V_1`

`T_2=(4.00color(red)cancel(color(black)("L"))xx293.2"K")/(6.00color(red)cancel(color(black)("L")))="195 K"`

The following website is very helpful:

http://chemistry.bd.psu.edu/jircitano/gases.html