If a sample of H2 has a volume of 22.4 L at 273.0 K and 760.0 mmHg, what will the volume be if the pressure decreases to 650.0 mmHg and the temperature increases to 320.0K?

30.7L

Explanation:

The following data were obtained from the question:

V1 (initial volume) = 22.4L

T1 (initial temperature) = 273K

P1 (initial pressure) = 760mmHg

P2 (final pressure) = 650mmHg

T2 (final temperature) = 320K

V2 (final volume) =?

Using the general gas equation P1V1/T1 = P2V2/T2, the final volume of the gas can be obtained as follow:

P1V1/T1 = P2V2/T2

760x22.4/273 = 650xV2/320

Cross multiply to express in linear form

273 x 650 x V2 = 760 x 22.4 x 320

Divide both side by 273 x 650

V2 = (760 x 22.4 x 320)/(273 x 650)

V2 = 30.7L

Therefore, the new volume of the gas will be 30.7L

the enthalpy of formation of water is given as -285.8kj/mol, which states that the enthalpy of formation is negative.

negative enthalpy of formation indicates that the heat is released during the reaction process. the heat is released from reaction to the surroundings the enthalpy is expressed with a negative sign with the energy given out in the reaction process.

the formation of water takes place by giving out heat which is -285.8 kj/mol.