An ideal gas, Cp= (5/2)R and Cv= (3/2)R is changed from P(initial)= 1bar and V(i

Question

An ideal gas, Cp= (5/2)R and Cv= (3/2)R is changed from P(initial)= 1bar and V(initial)= 12m^3 to P(final)= 12bar and V(final)= 1m^3 by the following mechanically                 reversible processes: (a) Isothermal compression. (b) Adiabatic compression followed by cooling at constant pressure. (c) Adiabatic compression followed by cooling                 at constant volume. (d) Heating at constant volume followed by cooling at constant pressure. (e) Cooling at constant pressure followed by heating at constant                 volume. Calculate Q, W, change in U, and change in H for each of these processes, and sketch the paths of all processes on a single PV diagram.

Answers: a) W=2982 KJ b) W=5106kJ c) W=7635kJ d) W=13200kJ e) W=1100kJ

I don't understand how to calculate a) since H and U require a change in T.

Explanation / Answer

PV=nRT

For isothermal process (a): Q = -W = RTln(V2/V1)

for isobaric processes: Q = ?H = ?Cp dT

Adiabatic Processes: TV^(?-1) = const, TP^(1-?)/? = const, PV^? = const,

for Isochoric processes: Q = ?U = ?Cv dT

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