Question

In a closed container, one gram of water (1 cm) is boiled under constant pressure of 1 atm (1.013 10Pa) until it changes into steam. The heat of vaporization at this pressure is Lv = 2.256 x 106J/kg = 2256 J/g. Assume that water vapor behaves like an ideal gas. The temperature of the steam is 373 K, number of moles of water vapor n = 0.0546 (g mol), the ideal gas constant R = 82.057 (cm3.atm/g mol-K). (a) How much volume (in cm) of steam is produced? (b) How much work is done by the water when it vaporizes? (c) How much has the internal energy increased?

Answer 1

of Pressuu Given data Initially Igm of water lie. 1cm 3 water is present. 1.013 X10S pa 3 latus Heat of vaporization tv = 2.9568100 J/kg = 2256 59 Temperature = 373 * mo. of moles of water = in 1 gram n = 0.0546 mol

R= 82.057 (cm3, atm, itm/gmolk Hena, from ideal gas equation- PV = nRT 0.0546 X 82.0578 373 MRT р 1 76 4.087m3 1671-156 cm3 b) work done = Pressur XAV AV = (1671.157 – 1) X Pressuse AV = 1670.156 X 1.013X105 N/m2 5 Nom -6 Av= 1670 .156 x 10°X 1.013 X10 17.264 kgfm AV- 169.187 Nm c) AHV Now, q= qe heat gained A Hwy heat of vaporization m- mass Hence, For 1 gram of water q= 1X 2256 Aty

q= 2256 J will increased by energy Hence, the internal tren 2256 Joule.