• # question_answer 540 calories of heat convert 1 cubic centimeter of water at ${{100}^{o}}C$ into 1671 cubic centimeter of steam at ${{100}^{o}}C$ at a pressure of one atmosphere. Then the work done against the atmospheric pressure is nearly A) 540 cal B) 40 cal C) Zero cal D) 500 cal

 [b] Amount of heat given $=540\ calories$ Change in volume $\Delta V=1670\ c.c$ Atmospheric pressure $P=1.01\times {{10}^{6}}\ dyne/c{{m}^{2}}$ Work done against atmospheric pressure $W=P\Delta V$$=\frac{1.01\times {{10}^{6}}\times 1670}{4.2\times {{10}^{7}}}\approx 40\ cal$