Question

42.9 liters of hydrogen are collected over water at 76.0 °C and has a pressure of 294 kPa. What would the pressure (in kPa) of the “dry” hydrogen at 38.7°C in a 22.8 liter container be?

Answer 1

494.1 kPa

Step-by-step explanation:

Using the combined gas law equation;

P1V1/T1 = P2V2/T2

Where;

P1 = initial pressure (kPa)

P2 = final pressure (kPa)

V1 = initial volume (L)

V2 = final volume (L)

T1 = initial temperature (K)

T2 = final temperature (K)

According to the information provided in this question,

P1 = 294 kPa

P2 = ?

V1 = 42.9 liters

V2 = 22.8 liters

T1 = 76.0°C = 76 + 273 = 349K

T2 = 38.7°C = 38.7 + 273 = 311.7K

294 × 42.9/349 = P2 × 22.8/311.7

12612.6/349 = 22.8 P2/311.7

36.14 = 22.8P2/311.7

Cross multiply

36.14 × 311.7 = 22.8P2

11264.605 = 22.8P2

P2 = 11264.605 ÷ 22.8

P2 = 494.1 kPa