A perfect gas undergoes isothermal compression which reduces it's volu...
Given
- Volume of gas before compression (V₁) = 1.80 dm³
- Final pressure of gas (P₂) = 1.48 × 10³ Torr
- Final volume of gas (V₂) = 2.14 dm³
Assumptions
- The gas is a perfect gas, which means it follows the ideal gas equation.
- The process is isothermal, which means the temperature remains constant throughout the process.
Solution
Step 1: Understanding the Problem
We have to find the original pressure of the gas before compression. To do this, we can use the ideal gas equation:
PV = nRT
Where:
- P is the pressure of the gas
- V is the volume of the gas
- n is the number of moles of the gas
- R is the ideal gas constant
- T is the temperature of the gas
In this problem, we can assume that the number of moles and the temperature of the gas remain constant. Therefore, the equation can be simplified to:
P₁V₁ = P₂V₂
Step 2: Calculating the Original Pressure in Torr
Substituting the given values into the equation, we have:
P₁(1.80 dm³) = (1.48 × 10³ Torr)(2.14 dm³)
Simplifying the equation, we get:
P₁ = (1.48 × 10³ Torr)(2.14 dm³) / 1.80 dm³
P₁ = 1.76 × 10³ Torr
Therefore, the original pressure of the gas before compression is 1.76 × 10³ Torr.
Step 3: Converting the Pressure to Bar
To convert the pressure from Torr to Bar, we use the conversion factor:
1 Bar = 750.06 Torr
Therefore, the original pressure in Bar is:
P₁ (Bar) = 1.76 × 10³ Torr / 750.06 Torr/Bar
P₁ (Bar) = 2.35 Bar
Therefore, the original pressure of the gas before compression is 2.35 Bar.