10 grams of molecular oxygen gas at an initial pressure of 2.5 atm and a tempera

Question

10 grams of molecular oxygen gas at an initial pressure of 2.5 atm and a temperature of 25?
undergoes an isochoric cooling until the pressure is halved.
(a) What is the temperature at the end of the process?
(b) How much work has been done in this step?
Next, the gas is isothermally compressed to its original pressure.
(c) What is the volume at the end of the compression?
(d) How much work has been done in this step?
Finally, the gas undergoes an isobaric expansion to its original volume.
(e) How much work has been done in this step?

Explanation / Answer

 PV=nRT   n=10/32

initial volume =nrt/p=3.05 L

u have not given the unit of tamperatue im taking that as celsius

so 25 c in KELVIN IS 298 K (273+C)

a)process is isochoric (volume contant)

   P1/T1=P2/T2

2.5/298=(2.5/2)/T2

T2=149 K

b) as the process is isochoric ie. dV=0  implies work done=P dV=0

C)now process is isothermal, ie. T CONSTANT

P1 V1=P2 V2

2.5/2 *3.05=2.5 *V2

1.525 L=V2

D)work done iin isothermal process=nRT ln v2/v1 =-268.33 J

 

e)isobaric ie P=CONTANT

v1/t1=v2/t2

1.525/149=3.25/t2

t2=298 k

back to initial temp ie 25 C

F) WORK DONE IN isobaric process=pdv=nR dT=387.12 J

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