Question

A balloon contains$2500\mathrm{mL}$of helium gas at $75°C$. What is the final volume, in milliliters, of the gas when the temperature changes to each of the following, if $P$and n do not change?

a. $55°C$

b.$680.K$

c. $-25°C$

d. $240.K$

Short answer

The final volume, in milliliters, of the gas when the temperature changes to each of the following, if P and n do not change :

a. $1833.3\mathrm{mL}$

b.$13566.6\mathrm{mL}$

c.$-832.5\mathrm{mL}$

d.$-1100\mathrm{mL}$

Step-by-step solution

Part(a) Step 1: Given information

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.

Part(a) Step 2: Explanation

By using charles law , we get ;

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

Part(b) Step 1: Given information

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.

Part(b) Step 2: Explanation

By using charles law , we get ;

We will convert the given temperature in celcius ,

$T_2=680-273=407°C$

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

Part(c) Step 1: Given information

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.

Part(c) Step 2: Explanation

By using charles law , we get ;

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

Part(d) Step 1: Given information

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.

Part(d) Step 2:Explanation

By using Charles law, we get ;

We will give temperature from Kelvin to Celcius

$T_2=240-273=-33°C$

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