Question

One kilogram of water fills a 150 L rigid container at an initial pressure of 2MPa. The container is cooled to 40 oC. Find the initial temperature and final pressure of the water.

Answer 1

Step-by-step explanation:

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Answer 2

The initial temperature is 649 K (376 °C).

The final pressure is 0.965 MPa

Step-by-step explanation:

From the ideal gas equation

PV = nRT

P is the initial pressure of water = 2 MPa = 2×10^6 Pa

V is intial volume = 150 L = 150/1000 = 0.15 m^3

n is the number of moles of water in the container = mass/MW = 1000 g/18 g/mol = 55.6 mol

R is gas constant = 8.314 m^3.Pa/mol.K

T (initial temperature) = PV/nR = (2×10^6 × 0.15)/(55.6 × 8.314) = 649 K = 649 - 273 = 376 °C

From pressure law,

P1/T1 = P2/T2

P2 (final pressure) = P1T2/T1

T2 (final temperature) = 40 °C = 40 + 273 = 313 K

P1 (initial pressure) = 2 MPa

T1 (initial temperature) = 649 K

P2 = 2 × 313/649 = 0.965 MPa