When a sample of neon with a volume of 683 mL and a pressure of 0.792 atm was heated from 19.0 to 76.0 °C, its volume became 733 mL. What was its final pressure (in atm)?

(P1V1/T1) = (P2V2/T2)

p2=.792atm x 683mL/ 292.15 x 349.15/733mL I got .882 and it says it wrong. Do you know what I done wrong?

I don't think you did anything wrong.

v1 = 683 mL
p1 = 0.792 atm
t1 = 19C = 273+19 = 292K

v2 = 733 mL
p2 = ? in atm.
t2 = 76C = 273+76 = 349K

(0.792*683/292) = (p2*733/349)

If I used 292.15 and 349.15 the answer is 0.8819 which rounds to 0.882 atm.

Usually when something like this happens it's because of the number of significant figures. Technically that 683 is really 700 (rounded to 1 s.f.) and the 733 is 700 rounded to 1 s.f. but I doubt that is the problem.

p2 = (0.792*683*349)/(292*733)
p2 = 0.882 atm.

To find the final pressure of the neon gas, we can use the combined gas law equation, which is derived from the ideal gas law. The combined gas law equation is given by:

(P1 × V1) / (T1) = (P2 × V2) / (T2)

where P1 and P2 are the initial and final pressures respectively, V1 and V2 are the initial and final volumes respectively, and T1 and T2 are the initial and final temperatures respectively.

Let's substitute the given values into the equation:

P1 = 0.792 atm
V1 = 683 mL = 0.683 L
T1 = 19.0 °C = 19.0 + 273.15 K (converting to Kelvin)

P2 = ?
V2 = 733 mL = 0.733 L
T2 = 76.0 °C = 76.0 + 273.15 K (converting to Kelvin)

Now, we can rearrange the equation to solve for P2:

P2 = (P1 × V1 × T2) / (V2 × T1)

Substituting the given values into the equation, we have:

P2 = (0.792 atm × 0.683 L × (76.0 + 273.15 K)) / (0.733 L × (19.0 + 273.15 K))

Calculating this expression will give us the final pressure:

P2 = (0.792 × 0.683 × (76.0 + 273.15)) / (0.733 × (19.0 + 273.15))

P2 ≈ 0.888 atm

Therefore, the final pressure of the neon gas is approximately 0.888 atm.