Question

A gas occupy a certain volume at stp what will be it's temperature if it's volume increase 3 times annd pressure is half?

Answer 1

`409.73"K"`, or `136.58^@"C"`

Explanation:

The conditions at `STP` are defined as:

• The pressure is `1"atm"`
• The temperature is `273.15"K"`
• The gas occupies a volume of `22.4"L"`.

We know that `PV=nRT`. This we can write as:

`(PV)/(nRT)=k`, and it follows that:

`(P_1V_1)/(nRT_1)=(P_2V_2)/(nRT_2)`

Which reduces to, as `n` is the same in both cases, and `R` is a constant:

`(P_1V_1)/T_1=(P_2V_2)/T_2`

We need to solve for `T_2`.

`P_1V_1T_2=P_2V_2T_1`

`T_2=(P_2V_2T_1)/(P_1V_1)`

We know that:

`V_1=22.4"L"`

`P_1=1"atm"`

`T_1=273.15"K"`

Since volume tripled and pressure halved, we have:

`P_2=0.5"atm"`

`V_2=67.2"L"`.

Inputting:

`T_2=(0.5*67.2*273.15)/(1*22.4)`

`T_2=409.73"K"`, or `136.58^@"C"`