A 1.50 L sample of neon gas at 1.10 atm and 25 °C is heated to 45 °C. The neon gas is then subjected to a pressure of 1.50 atm. Determine the new volume of the neon gas

1.17 L

Explanation:

Step 1: Given data

Step 2: Convert the temperatures to the Kelvin scale

We will use the following expression.

K = °C + 273.15

K = 25°C + 273.15 = 298 K

K = 45°C + 273.15 = 318 K

Step 3: Calculate the final volume of the neon gas

If we assume ideal behavior, we can calculate the final volume of the neon gas using the combined gas law.

P₁ × V₁ / T₁ = P₂ × V₂ / T₂

V₂ = P₁ × V₁ × T₂ / T₁ × P₂

V₂ = 1.10 atm × 1.50 L × 318 K / 298 K × 1.50 atm = 1.17 L

c.

explanation:

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 :