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This was a question from Pradeep book for Class 11 Physics. Is the question incomplete and requires more information? If no, please give me a solution. Thanks!

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1 Answer
Mar 9, 2018

Work done #W# in case of isobaric process is #nR delta T#

Given , #n=1# and #delta T=1 ^@C=1K# (as #1 ^@# Change in Celcius scale is equals to #1 K# change in Kelvin scale.)

And, #R=8.31 J K^-1 mol^-1#

So, #W=1*8.31*1=8.31J=8.31/4.2=2.08# Calories

Remember the following formulae for thermodynamics for quick answering,where #del Q# stands for heat energy required, #del U# stands for change in internal energy and #del W# stands for work done.

And in the given expressions, #V_2# is the final volume and #V_1# is the initial volume, #P_1# is initial pressure and #P_2# is final pressure,and #delta T# is change in temperature.

#gamma# stands for #C_p/C_v#

For Isothermal process

#del Q =nRT ln ((V_2)/(V_1))#

#del U =0#

#del W = del Q=nRT ln ((V_2)/(V_1))=nRT ln ((P_2)/(P_1))# (as #PV=constant#)

For Adiabatic process

#del Q=0#

#del U =nC_v delta T=(P_2 V_2 -P_1V_1)/(gamma -1)#

#del W=-del U=-nC_vdeltaT=(P_1 V_1 -P_2V_2)/(gamma -1)#

For isobaric process

#del Q=nC_pdeltaT#

#del U=nC_v deltaT#

#del W=nR deltaT#

For isometric process

#del Q=nC_vdeltaT#

#del U=del Q=nC_vdeltaT#

#del W=0#

Remember #T# is in Kelvin scale and don't forget to apply sign conventions

Also remember, #C_p=C_v+R# and #gamma =1 +2/f# where #f# is the degrees of freedom