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
