Answer:
0.01917 m^3/kg.
Explanation:
Given:
P = 15 MPa
= 1.5 × 10^4 kPa
T = 350 °C
= 350 + 273
= 623 K
Molar mass of water, m = (2 × 1) + 16
= 18 g/mol
= 0.018 kg/mol
R = 0.4615 kPa·m3/kg·K
Using ideal gas equation,
P × V = n × R × T
But n = mass/molar mass
V = (R × T)/P
V/M = (R × T)/P × m
= (0.4615 × 623)/1.5 × 10^4
= 0.01917 m^3/kg.
The specific volume of superheated water vapor under the conditions of 15 MPa pressure and 350°C temperature, using the ideal gas equation, is approximately 0.01919 cubic meter per kilogram.
The question is asking to calculate the specific volume of superheated water vapor using the ideal gas equation P = ρRT, where P is the pressure, ρ is the density (inverse of specific volume), R is the gas constant, and T is the temperature.
To find the specific volume (v), we need to rearrange the ideal gas equation to v = RT/P. Given that the pressure P = 15 MPa = 15000 kPa, the gas constant R = 0.4615 kPa.m³/kg.K, and the temperature T = 350°C = 623.15 K (adding 273 to convert °C to K), we can substitute these values into our rearranged equation balance to calculate for v.
The specific volume v = (0.4615 kPa.m³/kg.K * 623.15 K) / 15000 kPa = 0.01919 m³/kg. So, the specific volume of superheated water vapor under the given conditions is approximately 0.01919 cubic meter per kilogram.
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