A steel bottle contains 19.0 L of a gas at 11.0 atm and 22°C. What is the volume of gas at STP? Volume =

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**Problem Statement:**

A steel bottle contains 19.0 L of a gas at 11.0 atm and 22°C. What is the volume of gas at STP?

**Solution:**

To find the volume of the gas at standard temperature and pressure (STP), we use the ideal gas law and the combined gas law. 

- **STP Conditions:** 1 atm and 0°C (273 K)

**Given:**
- Initial Volume (V₁) = 19.0 L
- Initial Pressure (P₁) = 11.0 atm
- Initial Temperature (T₁) = 22°C = 295 K (Convert to Kelvin by adding 273)
  
**To Find:**
- Volume at STP (V₂) when P₂ = 1 atm and T₂ = 273 K

**Use the Combined Gas Law:**
\[
\frac{P₁ \times V₁}{T₁} = \frac{P₂ \times V₂}{T₂}
\]
Rearrange to solve for V₂:
\[
V₂ = \frac{P₁ \times V₁ \times T₂}{P₂ \times T₁}
\]

Substitute the known values:
\[
V₂ = \frac{11.0 \times 19.0 \times 273}{1 \times 295}
\]

Solve for V₂ to find the volume of gas at STP in liters.
Transcribed Image Text:**Problem Statement:** A steel bottle contains 19.0 L of a gas at 11.0 atm and 22°C. What is the volume of gas at STP? **Solution:** To find the volume of the gas at standard temperature and pressure (STP), we use the ideal gas law and the combined gas law. - **STP Conditions:** 1 atm and 0°C (273 K) **Given:** - Initial Volume (V₁) = 19.0 L - Initial Pressure (P₁) = 11.0 atm - Initial Temperature (T₁) = 22°C = 295 K (Convert to Kelvin by adding 273) **To Find:** - Volume at STP (V₂) when P₂ = 1 atm and T₂ = 273 K **Use the Combined Gas Law:** \[ \frac{P₁ \times V₁}{T₁} = \frac{P₂ \times V₂}{T₂} \] Rearrange to solve for V₂: \[ V₂ = \frac{P₁ \times V₁ \times T₂}{P₂ \times T₁} \] Substitute the known values: \[ V₂ = \frac{11.0 \times 19.0 \times 273}{1 \times 295} \] Solve for V₂ to find the volume of gas at STP in liters.
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