**-a) If the can is at $25 C$, what is the pressure in the can? ****-b) What volume would the propane occupy at STP? **

This **question** belongs to the chemistry **domain** and aims to **explain** the **concepts** of moles, **molar** mass, and **Ideal gas** equation. Furthermore, it **explains** how to calculate the **volume** and the **pressure** of the gas under any **given** condition.

**Molar mass** can be represented as “**mass** per mole.” It can also be **explained** as the sum of the mass of **all** the atoms in per **mole** of a substance. It is **expressed** in units of grams per **mole.** Molar mass is shown for **molecules** or **elements.** In the **matter** of single **molecules** or separate **atoms,** the molar **mass** would simply be the **element’s** mass described in **atomic** mass **units.** Similarly, the **atomic** mass and the molar **mass** of a **distinct** atom are **precisely** equal. As molar **mass** and atomic mass are **similar** for **individual** atoms, molar mass can be **used** to **estimate** the particle’s **uniqueness.**

Mole in **chemistry** is a formal scientific **unit** for counting **large** portions of very **small** commodities such as **molecules,** atoms, or other **defined** particles.

## Expert Answer

Given **Information:**

**Volume** = $250 mL$

Mass of the **propane** gas = $2.30g$

The **molar** mass of **propane** gas is given as = $44.1$

**Calculate** the number of **moles** of **propane** gas. The **formula** for finding **moles** is given as:

\[ Moles \space of \space propane = \dfrac{mass}{molar \space mass}\]

\[\dfrac{2.30g}{44.1}\]

\[Moles \space of \space propane = 00522mol \]

**Part A**

The given **temperature** is $25C$ which is $298K$.

The **pressure** can be **calculated** using the ideal **gas** equation:

\[PV=nRT\]

**Re-arranging** and making **pressure** $P$ the subject:

\[P=\dfrac{nRT}{V}\]

**Inserting** the values and **simplifying:**

\[=\dfrac{(0.0522)(0.0821) (298K) }{0.250}\]

\[=5.11 \space atm \]

**Part B**

The volume of **propane** occupied at STP can be **found** as follows:

\[P_1 = 1 \space atm\]

\[T_1 = 273 \space K\]

\[V_1 =?\]

Given **Conditions:**

\[P_1 = 5.11 \space atm\]

\[T_1 = 298 \space K\]

\[V_1 =0.250 \]

\[ \dfrac{P_1V_1}{T_1} = \dfrac{P_2V_2}{T_2} \]

**Inserting** the values:

\[ \dfrac{(1 \space atm)V_1}{273 \space K} = \dfrac{(5.11 \space atm) (0.250L)} {298 \space K} \]

\[ \dfrac{V_1}{273 \space K} = 0.00429 \]

\[ V_1 = 0.00429 \times 273 \space K \]

\[ V_1 = 1171 \space mL \]

## Numerical Answer

**Part A: **If the can is at $25 C$, the **pressure** in the can is $5.11 \space atm $.

**Part B: **The volume the **propane** occupies at **STP** is $1171 \space mL$.

## Example

They say that exposure to **temperatures** above $130 F$ may **cause** the can to **burst.** What is the **pressure** in the can at this **temperature?**

The **pressure** in the can at a **temperature** of $130 F$ is **found** as:

$130F$ **equals** to $327.4K$:

The **ideal** gas equation is **given** as:

\[PV = nRT \]

Re-arranging and **making** pressure $P$ the **subject:**

\[ P= \dfrac{nRT}{V} \]

**Inserting** the **values** and **simplifying:**

\[ = \dfrac{(0.0522) (0.0821) (327.4K) }{0.250} \]

\[ = 5.59 \space atm \]