Question

Problems and Exercises (a lot of these use partial derivatives!) Source: Heat and Thermodynamics, M. W. Zemansky and R. H. Dittman 1. A mass of mercury at STP is kept at constant volume. If the temperature is raised to 10°C, what will be the final pressure? From tables of physical constants, B and 2 of mercury remain practically constant within the above stated temperature range, B = 181 x 10'K' and x = 3.82 x 10" Pa' 2. A metal whose volume expansivity is 5 x 10'K' and isothermal compressibility is 1.2 x 10"Pa'is at a pressure of 10 x 10 Pa and a temperature of 20°C. A thick surrounding cover of invar, of negligible expansivity and compressibility, fits it very snugly. (a) What will be the final pressure if the temperature is raised to 32°C? (6) If the surrounding cover can withstand a maximum pressure of 1.2 x 10' Pa, what is the highest temperature to which the system can be raised? 3. A block of the same metal (see problem #2) is at a pressure of 1.0 x 10' Pa, a volume of 5 liters, and a temperature of 20°C undergoes a temperature rise of 12°C and an increase in volume of 0.5 cm. Calculate the final pressure.

Help with question number 3.

Answer 1

Given & Ans & initial pressure (P) = 1.0X105 pa initial volume (V) = 5 litresa = 5 x 103 cm3 initial Temperature (ti): 20°C - 20+273 = 293°4 increase in temperature (dt) = 12°C = 12+273 = 285 increase in volume (dv) = 0,5 cm3 volume expansivity (a) = 5x105 k- isothermal compressibility (B) = 1,2X10" Pal the change in pressure (dp) Related to volume expansivityla) & combustibility () can be expressed as formula' 1 dp = d.dT - L du i Idp- a do pv | the values in Egr (W & dp = 5x10-5(k=") x (285)(k) -[ 10.5) cm3 T 1.2860-** (Pa 1 .2x10"(fa") (5x10°cm3) | dp: 1187.5 X 106(Pa) – 0.0834 x 108(Pa) = 1187.5X100 – 8.34 X100 Pa dp = 1149.16 X80pa Hence the final pressure - do 1x105 pa + 1179.16X106 pa Petit ap 0.5X106 + 1179.16 X 106 po Pa = 1179.26 X106 Pa anned with scanner Pa = 1:17926 X109 Pal Scanner WE