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, the sign of the Joule-Tompson coefficient can be the equation of state of the real gas
predicted from
in the van der Waals equation of state, P=(nRT/(V-nb))- molecular diameter increases
((n^2a)/V^2)), the terms, nb, will increase as the
under what condition is H2 in the state corresponding 33 K, .33 atm
to N2 at 126 K and 1 atm?
the isothermal compressibility kT = -(1/V)(dV/dP)T, for (RT/P^2)/(RT/P +b)
the hard sphere equation of state P(V-nb) = nRT is given
by
definition of reduce temperature Tr=T/Tc
reduced pressure Pt=P/Pc
definition of the corresponding states it must have the same reduced temperature and pressure
the valve between the 2.00 L bulb, in which the gas After the valve is opened, the total volume is (2.00 + 3.00) L + 5.00 L and the
pressure is 1.00 atm, and the 3.00 L bulb, in which the total number of moles is the same of moles initially in separate bulbs. Since
gas pressure is 1.50 atm, is opened. What is the final both, the volume and number of moles add, 1.00 atm2.00 L + 1.50 atm3.00 L =
pressure in the two bulbs, the temperature being P*5.00 L so
constant and the same in both bulbs? P=1.30 atm
ideal gas equation of state at constant temperature the directly proportional to the product of pressure and volume, PV
number of moles is
what kind of isotherms graph is experimentally negative wave like exponential graph
obsessed near the critical temperature of the real gas? This shows P - V behavior of real gas near the critical point. The critical point
itself occurs when (∂P/∂V)T = 0 and (∂^2P/∂V^2)T=0
as pressure and temperature are increased to the -Δ(vaporization)H goes to 0
critical point... -the density of the fas approaches the same value as that of the liquid
-the index of refraction of the gas approaches the same value as that of the
liquid
the van der Waals equation of state (P+(n^2a/V^2))(V- 1 cm^3/mol
nb)=nRT contains a term representing a "molecular size".
The approximate magnitude of this term is
terms in real gas equations of state and their V=4/3pir^3
approximate magnitude r=10^-8 cm/molecule
V=(410^-24)(610^23)=24*10^-1 = 1 cm^3/mol