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