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a heating coil is immersed in 100 gram of sample of water at 1 atm and hundred degree celsius in a closed vessel in this heating process 60% of the liquid is converted to the gaseous form at constant pressure of 1 atm the density of liquid and gases water and at this condition of 1000kg/m^3 and 0.6kg/m^3 respectively find the magnitude of the work done

a heating coil is immersed in 100 gram of sample of water at 1 atm and hundred degree celsius in a closed vessel in this heating process 60% of the liquid is converted to the gaseous form at constant pressure of 1 atm the density of liquid and gases water and at this condition of 1000kg/m^3 and 0.6kg/m^3 respectively find the magnitude of the work done

Grade:12

1 Answers

Naveen
15 Points
4 years ago
Initially:
    100g sample of H2o(liquid)
 Density=Mass/volume
 Volume=Mass/density
 Mass=100/1000=0.1
 Density= 1000
So, volume=0.1/1000=10-4
Now,
60g gas + 40g water
Volume=V1+V2=M1/D1 +M2+D2
              =(60/1000)/0.6 +(40/1000)/1000
              =10-1+4×10-5
Work done=P(∆V)
                    =1×(V2-V1)
                     =(10-1+4×10-5)-(10-4)
                     =0.09994
                     =0.09994×1000 Liter-atm/K mole
                     =99.94×100 joule/K mole
                     =9994 J/K mol
 
                   
 

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