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A perfect gas undergoes isothermal compression which reduces it's volume by 1.80dm³. The final pressure and volume of gases are 1.48×10³ Torr and 2.14dm³ respectively; calculate the original pressure in (i) Torr and (ii) Bar?
Verified Answer
A perfect gas undergoes isothermal compression which reduces it's volu...
1 torr = 0.00133322 bar
148 x 1000 Torr = 197.317105 bar ~ 197 bar

This question is part of UPSC exam. View all NEET courses
Most Upvoted Answer
A perfect gas undergoes isothermal compression which reduces it's volu...
1 torr = 0.00133322 bar
148 x 1000 Torr = 197.317105 bar ~ 197 bar
Community Answer
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.
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A perfect gas undergoes isothermal compression which reduces it's volume by 1.80dm³. The final pressure and volume of gases are 1.48×10³ Torr and 2.14dm³ respectively; calculate the original pressure in (i) Torr and (ii) Bar?
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A perfect gas undergoes isothermal compression which reduces it's volume by 1.80dm³. The final pressure and volume of gases are 1.48×10³ Torr and 2.14dm³ respectively; calculate the original pressure in (i) Torr and (ii) Bar? for NEET 2024 is part of NEET preparation. The Question and answers have been prepared according to the NEET exam syllabus. Information about A perfect gas undergoes isothermal compression which reduces it's volume by 1.80dm³. The final pressure and volume of gases are 1.48×10³ Torr and 2.14dm³ respectively; calculate the original pressure in (i) Torr and (ii) Bar? covers all topics & solutions for NEET 2024 Exam. Find important definitions, questions, meanings, examples, exercises and tests below for A perfect gas undergoes isothermal compression which reduces it's volume by 1.80dm³. The final pressure and volume of gases are 1.48×10³ Torr and 2.14dm³ respectively; calculate the original pressure in (i) Torr and (ii) Bar?.
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