Question

An air bubble has a volume of $0.500L$at $18°C$ What is the final volume, in liters, of the gas when the temperature changes to each of the following, if $P$ and $n$ do not change?
a. $0°C$
b. $425K$
c. $-12°C$
d. $575K$

Short answer

Part(a) The final volume is $0.469L$

Part(b) The final volume is$0.730L$

Part(c) The final volume is $0.448L$

Part(d) The final volume is $0.988L$

Step-by-step solution

Part(a) Step 1 : Given information

We are given an air bubble with constant pressure and number of moles. We need to find final volume.

Part(a) Step 2 : Simplify

As we know, $0°C=273K$

Using Charle's Law $\frac{V_1}{T_1}=\frac{V_2}{T_2}$

$$\begin{gathered} V_1=0.500L \\ {T}_1=18°C=18+273K=291K \\ {T}_2=0°C=0+273K=273K \end{gathered}$$

Substituting the values, we get Final volume is $0.469L$

Part(b) Step 1 : Given information

We are given an air bubble with constant pressure and number of moles. We need to find final volume.

Part(b) Step 2 : Simplify

As we know, $0°C=273K$

Using Charle's Law $\frac{V_1}{T_1}=\frac{V_2}{T_2}$

$$\begin{gathered} V_1=0.500L \\ {T}_1=18°C=18+273K=291K \\ {T}_2=425K \end{gathered}$$

Substituting the values, we get Final volume is $0.730L$

Part(c) Step 1 : Given information

We are given an air bubble with constant pressure and number of moles. We need to find final volume.

Part(c) Step 2 : Simplify

As we know $0°C=273K$

Using Charle's Law $\frac{V_1}{T_1}=\frac{V_2}{T_2}$

$$\begin{gathered} V_1=0.500L \\ {T}_1=18°C=18+273K=291K \\ {T}_2=-12°C=-12+273K=261K \end{gathered}$$

Substituting the values, we get Final volume is $0.448L$

Part(d) Step 1 : Given information

We are given an air bubble with constant pressure and number of moles. We need to find final volume.

Part(d) Step 2 : Simplify

As we know, $0°C=273K$

Using Charle's Law $\frac{V_1}{T_1}=\frac{V_2}{T_2}$

$$\begin{gathered} V_1=0.500L \\ {T}_1=18°C=18+273K=291K \\ {T}_2=575K \end{gathered}$$

Substituting the values, we get Final volume is $0.988L$