Question

A student collects `450` `mL` of `HCl (g)` at a pressure of `100` `kPa` and a temperature of `17°C`. What is the new volume of the `HCl` at standard temperature and pressure?

Answer 1

Take `V_1/T_1=V_2/T_2` and rearrange to:

`V_2=(V_1T_2)/T_1=(0.450xx273)/290=0.424` `L` = `424` `mL`.

Explanation:

First convert the temperature to kelvin: `17+273=290` `K`. Standard temperature is `273` `K`.

Now `100` `kPa` is already standard pressure, so the pressure is constant.

We should work in SI units, so `450` `mL` = `0.450` `L`.

From the Combined Gas Law, but with pressure omitted because it is constant, take `V_1/T_1=V_2/T_2` and rearrange to `V_2=(V_1T_2)/T_1=(0.450xx273)/290=0.424` `L` = `424` `mL`.