A sample of an ideal gas has a volume of 2.23 L at 289 K and 1.05 atm. Calculate the pressure when the volume is 1.08 L and the temperature is 304 K. P= atm

2.28 atm

Explanation:

V₁ = 2.33L,  V₂ = 1.08L

T₁ = 289K,  T₂ = 304K

P₁ = 1.05 atm, P₂ = ?

Where V₁ and V₂ are initial and final volume respectively

T₁ and T₂ are initial and final temperature respectively

P₁ and P₂ are initial and final pressure respectively

The formula to be used here is the general gas equation:

P₁V₁/T₁=P₂V₂/T₂

1.05 × 2.23/289 = P₂ × 1.08/304

P₂ × 1.08 × 289 = 1.05 × 2.23 × 304

P₂ = (1.05 × 2.23 ×304) ÷ (1.08 × 289)

P₂ = 711.82 ÷ 312.12

P₂ = 2.28 atm

answer

i think the answer is a) as air pressure in an area increases, the density of the gas particles in that area decreases.

hope it

a polar bond is formed when electrons are shared unevenly between atoms