A certain mass of gas occupies 5.5litres at 300k and 650torr. What will be its volume in litres if it is cooled to 283k and its pressure is increased to 980torr?

Question

Answer

Answer

Given Information:
- Initial volume (V1) = 5.5 liters
- Initial temperature (T1) = 300 K
- Initial pressure (P1) = 650 torr
- Final temperature (T2) = 283 K
- Final pressure (P2) = 980 torr

Boyle's Law:
Boyle's Law states that at constant temperature, the volume of a given mass of gas is inversely proportional to its pressure. Mathematically, it can be represented as P1V1 = P2V2.

Solution:
To find the final volume (V2), we can rearrange Boyle's Law equation as follows:
V2 = (P1 * V1) / P2

Step 1: Convert initial and final temperatures to Kelvin
Initial temperature (T1) = 300 K
Final temperature (T2) = 283 K

Step 2: Convert initial and final pressures to atmospheres
Initial pressure (P1) = 650 torr / 760 torr/atm = 0.855 atm
Final pressure (P2) = 980 torr / 760 torr/atm = 1.289 atm

Step 3: Substitute the values into the formula and solve for V2
V2 = (0.855 atm * 5.5 liters) / 1.289 atm
V2 ≈ 3.635 liters

Conclusion:
The volume of the gas will be approximately 3.635 liters when it is cooled to 283 K and its pressure is increased to 980 torr.

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