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I. There are currently three different ways of defining the standard state of a gas: 1. A temperature of 273.15 K (0°C) and a pressure of 101.325 kPa. The standard state defined by the agreement of the 10th General Conference on Weights and Measures (CGPM) in 1954 is: a temperature of 273.15 K (0°C) and a pressure of 101.325 kPa. This standard state is widely adopted in the field of science and technology around the world. SH3005-1999 \"Code for Selection and Design of Automation Instruments in Petrochemical Industries\" also adopts this standard state (Article 5.1.1). 2. Temperature: 288.15 K (15°C), pressure: 101.325 kPa. The International Organization for Standardization and American **standards specify 288.15 K (15°C) as temperature and 101.325 kPa as pressure to serve as the standard conditions for measuring the volumetric flow rate of gases. 3. Temperature: 293.15 K (20°C), pressure: 101.325 kPa. GB/T21446-2008 \"Measurement of natural gas flow rate using standard orifice plate flow meters\" and SY/T6143-2004 \"Standard orifice plate measurement method for natural gas flow rate\" adopt this standard state. II. Volume conversion between the standard state and the actual state of a gas: Based on the ideal gas law PV=nRT, we obtain PV/T=nR. For a fixed amount of gas, (P1 V1)/T1 = (P2 V2)/T2, where P1 is the pressure under standard conditions, in MPa ； V1 is the volume under standard conditions, in Nm3/h as the unit ； T1 is the temperature under standard conditions, unit: K ； P2 is the pressure under actual conditions, unit: MPa ； V2 represents the volume under actual conditions, in units of m3/h ； T2 is the temperature under actual conditions, in units of K. III. Example: If a certain gas has a volume of 1050 m3/h at 40°C and 0.95 MPa, what is its volume under standard conditions (20°C, 101.325 kPa)? According to (P1 V1)/T1 = (P2 V2)/T2, it follows that (1.01325×V1)/(273.15+20) = (0.95×1050)/(273.15+40); therefore, V1 = 922 Nm3/h.