Question

A sample of methane has a volume of $25$mL at a pressure of $0.80$ atm. What is the final volume, in milliliters, of the gas at each of the following pressures, if there is no change in temperature and amount of gas?
a.$0.40$atm

b. $2.00$ atm
c. $2500$mmHg

d. $80.0$torr

Short answer

Part(a) The final volume is $50mL$

Part(b) The final volume is $10mL$

Part(c) The final volume is $6.08mL$

Part(d) The final volume is $190mL$

Step-by-step solution

Part(a) Step 1 : Given information

We need to find the final volume of gas for given data.

Part(a) Step 2: Explanation

Using Boyle's law

$P_1{V}_1=P_2{V}_2$

Now,

$$\begin{gathered} {\mathrm{P}}_1=0.80\mathrm{atm} \\ {\mathrm{V}}_1=25\mathrm{mL} \\ {\mathrm{P}}_2=0.40\mathrm{atm} \end{gathered}$$

Substituting the given values we get,

$$\begin{gathered} 0.8\times 25=0.40\times {\mathrm{V}}_2 \\ {V}_2=50\mathrm{mL} \end{gathered}$$

Part(b) Step 1 : Given information

We need to find the final volume of gas for given data.

Part(b) Step 2: Explanation

Using Boyle's law

$P_1{V}_1=P_2{V}_2$

Now,

$$\begin{gathered} {\mathrm{P}}_1=0.80\mathrm{atm} \\ {\mathrm{V}}_1=25\mathrm{mL} \\ {\mathrm{P}}_2=2.00\mathrm{atm} \end{gathered}$$

Substituting the given values we get,

$$\begin{gathered} 0.80\times 25=2\times {\mathrm{V}}_2 \\ {\mathrm{V}}_2=10\mathrm{mL} \end{gathered}$$

Part(c) Step 1 : Given information

We need to find the final volume of gas for given data.

Part(c) Step 2: Explanation

Using Boyle's Law

$P_1{V}_1=P_2{V}_2$

Now,

$$\begin{gathered} {\mathrm{P}}_1=0.80\mathrm{atm}=608\mathrm{mmHg} \\ {\mathrm{V}}_1=25\mathrm{mL} \\ {\mathrm{P}}_2=2500\mathrm{mmHg} \end{gathered}$$

Substituting the values we get

$$\begin{gathered} 608\times 25=2500\times {\mathrm{V}}_2 \\ {\mathrm{V}}_2=6.08\mathrm{mL} \end{gathered}$$

Part(d) Step 1 : Given information

We need to find the final volume of gas for given data.

Part(d) Step 2: Explanation

Using Boyle's Law

$P_1{V}_1=P_2{V}_2$

Now,

$$\begin{gathered} {\mathrm{P}}_1=0.80\mathrm{atm}=608\mathrm{torr} \\ {\mathrm{V}}_1=25\mathrm{mL} \\ {\mathrm{P}}_2=80\mathrm{torr} \end{gathered}$$

substituting the values, we get

$$\begin{gathered} 608\times 25=80\times {\mathrm{V}}_2 \\ {\mathrm{V}}_2=190\mathrm{mL} \end{gathered}$$