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The standard state of a material (pure substance, mixture or solution) is a reference point used to calculate its properties under different conditions.A degree sign (°) or a superscript Plimsoll symbol (⦵) is used to designate a thermodynamic quantity in the standard state, such as change in enthalpy (ΔH°), change in entropy (ΔS°), or change in Gibbs free energy (ΔG°).
Since 1982, STP has been defined as a temperature of 273.15 K (0 °C, 32 °F) and an absolute pressure of exactly 10 5 Pa (100 kPa, 1 bar). NIST uses a temperature of 20 °C (293.15 K, 68 °F) and an absolute pressure of 1 atm (14.696 psi, 101.325 kPa). [3] This standard is also called normal temperature and pressure (abbreviated as NTP).
The ideal gas equation can be rearranged to give an expression for the molar volume of an ideal gas: = = Hence, for a given temperature and pressure, the molar volume is the same for all ideal gases and is based on the gas constant: R = 8.314 462 618 153 24 m 3 ⋅Pa⋅K −1 ⋅mol −1, or about 8.205 736 608 095 96 × 10 −5 m 3 ⋅atm⋅K ...
Only certain lighter than air gases are suitable as lifting gases. Dry air has a density of about 1.29 g/L (gram per liter) at standard conditions for temperature and pressure (STP) and an average molecular mass of 28.97 g/mol, [1] and so lighter-than-air gases have a density lower than this.
Gay-Lussac used the formula acquired from ΔV/V = αΔT to define the rate of expansion α for gases. For air, he found a relative expansion ΔV/V = 37.50% and obtained a value of α = 37.50%/100 °C = 1/266.66 °C which indicated that the value of absolute zero was approximately 266.66 °C below 0 °C. [ 12 ]
Until 1982, STP was defined as a temperature of 273.15 K (0 °C, 32 °F) and an absolute pressure of 101.325 kPa (1 atm). Since 1982, STP is defined as a temperature of 273.15 K (0 °C, 32 °F) and an absolute pressure of 100 kPa (1 bar). Conversions between each volume flow metric are calculated using the following formulas: Prior to 1982,
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For a substance X with a specific volume of 0.657 cm 3 /g and a substance Y with a specific volume 0.374 cm 3 /g, the density of each substance can be found by taking the inverse of the specific volume; therefore, substance X has a density of 1.522 g/cm 3 and substance Y has a density of 2.673 g/cm 3. With this information, the specific ...