Calculate the work done when 1.000 mole of a perfect gas expands reversibly from 1.0 L to 10 L at 298.0 K. Then, calculate the amount of work done when the gas expands irreversibly against a constant external pressure of 1.00 atm. Compare the two values and comment.

See explanation

Explanation:

For work done at constant temperature;

w = -nRTln(V2/V1)

V1 = 1.0 L

V2 = 10 L

w = -(1 * 8.314 * 298) ln(10/1)

w = -5704.8 J

When the gas expands irreversibly against a constant external pressure;

w = -(PΔV)

w = -(1.00 (10 - 1))

w = - 9 J

During expansion of a gas in which the initial and final limits of volume are given, the constant pressure process produces more work than the isothermal process as we have seen above.

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.