A sample of 0.165-mol perfect gas occupies 20.0 dm3 at 310 K initially. Process (1) The gas expands isothermally against a constant external pressure of  8.5 kPa until its volume becomes 50.0 dm3. Process (2) The gas expands isothermally and reversibly from 20.0 dm3 to 50.0 dm3 at 310 K. Which of the above processes does have a greater heat transaction (namely q)?       Process (1)       Process (2)         Insufficient information.     Same for Process (1) and Process (2)

For both circumstances (reversible and irreversible) what is the change in enthalpy dS for the gas, the surroundings (assuming ideal isothermal bath), and the universe?

Suppose that attractions are the dominant interaction between gas molecules, and the equation of state is p = nRT/V – n2a/V2. Determine the work (W(non-ideal gas)) of reversible, isothermal expansion of such a gas from initial volume V (initial) = 20.0 L to final volume V(final) = 40.0 L if n = 2.00 mol, T = 300 K, and a = 3.621 atm-L2/mol2. Watch your units. Determine the work (W(ideal gas) of reversible, isothermal expansion of an ideal gas from initial volume V (initial) = 20.0 L to final volume V(final) = 40.0 L if n = 2.00 mol and T = 300 K. Show the difference W(non-ideal) – W(ideal). If all your calculations are done correctly, this result shows you the effect of attractive interaction between gas particles on the work done by the system.

Everything needed is included Consider the reversible and isothermal compression of 2.0 moles of ammonia (NH3 (g)) from an initial volume of 10.0 L to a final volume of 1.00 L, at a constant temperature of 298 K: 6a. Calculate the work w and heat q associated with this change in volume, considering NH3 to be an ideal gas. 6b. Calculate the work w (but not the heat) using the virial equation of state, with second virial coefficient B = −0.297 L/mol for NH3 at this temperature (and ignoring any higher coefficients).