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

in scientific notation, the value of the  speed of light in a vacuum

  will be written as 2.995 * 10^8. when this value is represented in the form of

  a * 10^b, then, a = 2.995 and b = 8.

the value of the speed of light in a vacuum given above can also be rounded up to approximately 300000000. in this case, the scientific notation will be written as 3 *10^8.

answer: the value of enthalpy change will be -152.80 kj.

solution:

pressure = 1atm , temperature = = 298 k

given that pressure and temperature are same in both gases , the number of moles of both gases can be determined by the ideal gas equation.

according to reaction 3 moles no are reacting with 1 mol of ,then 0.49 mol of no will react with moles of ozone that is 0.16 moles of ozone. since, no is limiting reagent and is an excessive reagent.

enthalpy change for the reaction between the given volumes of the gases is :