At roughly what pressure, P, and temperature, T, willCarbon phase diagramdiamond, graphite, and liquid carbon all exist in equilibrium?106–Solid I (diamond)10°–LiquidP =atm10ª–103-Solid II(graphite)10²-GasT =K10-1-100020003000400050006000T (K)Which solid form of carbon is more dense?diamondO graphiteP (bar)

From the above phase diagram, the roughly estimated pressure and temperature of carbon compounds i.e. graphite, diamond and liquid carbon all exist in equilibrium are approximately 105 bar and 4300 K respectively. As we know 1bar = 0.986atm So pressure = (0.986×105 )atm = 98692.3atm which is equivalent to 105atm.As we know pressure is directly proportional to density. From the above graph, diamond is above graphite i.e. diamond exists at higher pressure, making it more denser than graphite.

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Chemistry

At roughly what pressure, P, and temperature, T, will Carbon phase diagram diamond, graphite, and liquid carbon all exist in equilibrium? 10- Solid I (diamond) 10- Liquid P = atm 10- 10- Solid II 10- (graphite) Gas T = K 10- 1- 1000 2000 3000 4000 5000 6000 T (K) Which solid form of carbon is more dense? O diamond O graphite P (bar)

At roughly what pressure, P, and temperature, T, will Carbon phase diagram diamond, graphite, and liquid carbon all exist in equilibrium? 10- Solid I (diamond) 10- Liquid P = atm 10- 10- Solid II 10- (graphite) Gas T = K 10- 1- 1000 2000 3000 4000 5000 6000 T (K) Which solid form of carbon is more dense? O diamond O graphite P (bar)

Chemistry

10th Edition

ISBN:9781305957404

Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Publisher:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCoste

Chapter1: Chemical Foundations

Section: Chapter Questions

Problem 1RQ: Define and explain the differences between the following terms. a. law and theory b. theory and...

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