Question

Solve the above questions please

Answer 1

The solution to the lower question yields the answer 293 cal/K by following the stated method -
Approach the question using these relations -
∆Q = m*Li , where m is mass and Li is latent heat.
dS=dQ/T , is the relation for entropy change.
Now , ∆Q = 1000 * 80 = 80000
Temperature is given as 0℃ , ie 273K.
Putting the values in entropy relation ,
dS = 80000/273 = 293.04 or 293 (approx.)

As for the upper question , there seems to be something missing from the methods I used and that's why I'm getting the answer 1-(V2/V1)^gamma but the tick mark shows the answer to be 1-(V2/V1)^(1-gamma). Approach the problem with the relation PV^gamma is constant in adiabatic relation. See if you get a workaround do tell me or if I get something I'll add it here.