91Ó°ÊÓ

We need to find that final volume, in milliliters, of the gas when the temperature changes to each of the following, if Pressure and amount do not change.

By using charles law , we get ;

${\mathrm{V}}_2=\frac{{\mathrm{V}}_1×{\mathrm{T}}_2}{{\mathrm{T}}_1}=\frac{2500×55}{75}=1833.3\mathrm{mL}$

We need to find that final volume, in milliliters, of the gas when the temperature changes to each of the following, if Pressure and amount do not change.

By using charles law , we get ;

We will convert the given temperature in celcius ,

$T_2=680-273=407°C$

localid="1654148134138" ${\mathrm{V}}_2=\frac{{\mathrm{V}}_1×{\mathrm{T}}_2}{{\mathrm{T}}_1}=\frac{2500×407}{75}=13566.6\mathrm{mL}$

We need to find that final volume, in milliliters, of the gas when the temperature changes to each of the following, if Pressure and amount do not change.

By using charles law , we get ;

$$\begin{gathered} {\mathrm{V}}_2=\frac{{\mathrm{V}}_1×{\mathrm{T}}_2}{{\mathrm{T}}_1}=\frac{2500×\left(-25\right)}{75} \\ =-832.5\mathrm{mL} \end{gathered}$$

We need to find that final volume, in milliliters, of the gas when the temperature changes to each of the following, if Pressure and amount do not change.

By using Charles law, we get ;

We will give temperature from Kelvin to Celcius

$T_2=240-273=-33°C$

localid="1654148240898" ${\mathrm{V}}_2=\frac{{\mathrm{V}}_1×{\mathrm{T}}_2}{{\mathrm{T}}_1}=\frac{2500×\left(-33\right)}{75}=-1100\mathrm{mL}$