Question

A gas has a pressure of 699.0 mm Hg at 40.0 °C. What is the temperature at standard pressure?

Answer 1

To solve this problem, we can use the combined gas law, which relates the pressure, volume, and temperature of a gas:

(P1 × V1) ÷ T1 = (P2 × V2) ÷ T2

where P1 and T1 are the initial pressure and temperature, P2 is the final pressure (in this case, standard pressure, which is 760 mm Hg), V1 and V2 are the initial and final volumes (which we can assume to be constant), and T2 is the final temperature that we want to find.

We can rearrange this equation to solve for T2:

T2 = (P2 × V1 × T1) ÷ (P1 × V2)

Substituting the given values, we get:

T2 = (760 mmHg × 1 L × 313 K) ÷ (699.0 mmHg × 1 L)

T2 ≈ 342 K

Therefore, the temperature at standard pressure is approximately 342 K or 69 °C.

Answer 2

336 K.

Make sure when solving to convert the Celsius temperature to Kelvin temperature by adding 273.15 to the Celsius temperature. :)