Question: Below is a PV diagram showing a cyclic process for 0.0040 moles of an

Question

                    Question: Below is a PV diagram showing a cyclic process for
                    0.0040 moles of an ideal monatomic gas. The direction of the cycle is shown by the arrows on the curve. The temperature does NOT change along                     segment ca

Below is a PV diagram showing a cyclic process for 0.0040 moles of an ideal monatomic gas. The direction of the cycle is shown by the arrows on the curve. The temperature does NOT change along segment ca - the process is isothermal from point c to point a. The other 2 sections of curve are graphically vertical and horizontal. What volume does the gas occupy at point a? What is the temperature at points a, b and c? For parts (c), (d) and (e) below, indicate whether the heat Q has gone into or out of the gas by referring to the sign, positive or negative, of your answer. How much heat went into or out of the gas during segment ab? How much heat went into or out of the gas during segment bc? How much heat went into or out of the gas during segment ca? For parts (f) , (g) and (h) below, indicate whether the change in internal energy ?E int is positive or negative by referring to the sign of your answer. Find the change in internal energy during segment ab. Find the change in internal energy during segment bc. Find the change in internal energy during segment ca.

Explanation / Answer

a) using the ideal gas equation ,

Pressure*Volume=constant (for the position a and c),

2*0.2=0.5*Va

or Va=0.8 *10^-3 m^3

so volume at a=0.8*10^-3 m^3

b)using the ideal gas equation,

PV=nRT,

for a,

0.5*0.8=0.004*(T/12)

or Ta=1200 K

For B,

2*0.8=0.004*(T/12)

or Tb=4800 K

for C,

2*0.2=0.004*(T/12)

or Tc=1200 K

c) Heat GONE into the system for the phase a to b,

Q ab=nCv dT

=0.004*(1.5*8.31)*(4800-1200)

=179.496 J

d)Heat FLOWN OUT of the system for b to c,

Qbc=n Cp dT

= 0.004*(2.5*8.31)*(1200-4800)

= -299.16 J

e)heat gone into the system= work done

=nRT ln(Vf/Vi)

=0.004*8.31*1200*ln(0.8/0.2)

=55.2965 J

f)Uab= n Cv dT

=0.004*1.5*8.31*(4800-1200)

=179.496 J

g) Ubc= n Cv dT

=0.004*1.5*8.31*(1200-4800)

= -179.496 J

c) Uca= nCv dT

but since this is an isothermal change,

temperature is constant and hence the inter energy change is 0

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