Question #f4956

1 Answer
Mar 3, 2017

By First law of thermodynamics we have

Q=dU+W.....[1]

Where
Q-> "heat given to the system" =nC_pdT

dU-> "change in internal energy of the system" =nC_vdT

W-> "work done by the system"

Here n-> "number of mole of the gas"

C_p->" molar heat capacity of the gas at constant pressure"

C_v->" molar heat capacity of the gas at constant volume"

dT-> "change in temprature"

So equation [1] becomes

nC_pdT=nC_vdT+W

=>W=n(C_p-C_v)dT=nRdT

Inserting

n=1mol

R =0.082LatmK^-1mol^-1=0.0082kJK^-1mol^-1

(as 1Latm =-.1kJ)

dT=(273-373)K=-100K

So work done by the system

=>W=nRdT=1molxx0.0082kJK^-1mol^-1xx(-100K)

=-0.82kJ

Here work done by the system is negative . So this means that work is to be done on the system for the cooling process as the process is not adiabatic one and the amount of work done on the system will be 0.82kJ