A gas sample is held at constant pressure. The gas occupies 2.97 L of volume when the temperature is 21.6°C. Determine the temperature at which the volume of the gas is 3.42 L. (Charles')

339.2K

Explanation:

Using Charles law equation;

V1/T1 = V2/T2

Where;

V1 = initial volume (L)

V2 = final volume (L)

T1 = initial temperature (K)

T2 = final temperature (K)

According to the information provided in this question,

V1 = 2.97 L

V2 = 3.42 L

T1 = 21.6°C = 21.6 + 273 = 294.6K

T2 = ?

Using V1/T1 = V2/T2

2.97/294.6 = 3.42/T2

Cross multiply

2.97 × T2 = 294.6 × 3.42

2.97T2 = 1007.532

T2 = 1007.532 ÷ 2.97

T2 = 339.236

The final temperature is 339.2K

n+h→nh3

we must remember that nitrogen and hydrogen are both diatomic molecules in their standard gas form. this adjusts the equation to

n2+h2→nh3

now we need to adjust coefficients in order to balance the atoms on each side of the equation. currently we have 2 atoms of nitrogen and 2 atoms of hydrogen on the reactant side and 1 atom of nitrogen and 3 atoms of hydrogen on the product side.

we can balance the hydrogens by placing a coefficient of 2 in front of ammonia and a coefficient of 3 in front of the hydrogen.

n2+3h2→2nh3

this gives us 6 hydrogen on each side and coincidentally the nitrogens now equal 2 on each side.