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Question: How do you calculate standard temperature and pressure?...

How do you calculate standard temperature and pressure?

Explanation

Solution

Standard temperature and pressure are standard sets of conditions for experimental measurements to be established to permit comparisons to be made between different sets of knowledge. The standard temperature and pressure (STP)\left( {STP} \right) are defined as 0C{{\text{0}}^ \circ }C and 1atm1atm of pressure.

Complete step by step solution:
Standard temperature and pressure describe the environment of a reaction. The quality temperature is adequate to273.15 K =0C=32F273.15{\text{ }}K{\text{ }} = 0^\circ C = 32^\circ F . The quality pressure equals to1 atm=760 Torr=760 torr = 101.35kPa.1{\text{ }}atm = 760{\text{ }}Torr = 760{\text{ }}torr{\text{ }} = {\text{ }}101.35kPa.
The ideal gas equation is written as follows:
PV=nRTPV = nRT
The above equation is a superb tool for understanding the connection between the quantity , pressure, temperature and amount of a perfect gas during a defined environment which will be controlled for a continuing volume.
By rearranging the above equation, we will find pressure using the subsequent equation:
P = nRTVP{\text{ }} = {\text{ }}\dfrac{{nRT}}{V}
STPSTP in chemistry is the abbreviation for normal Temperature and Pressure. STPSTP most ordinarily is employed when performing calculations on gases, like gas density. The quality temperature is 273 K (0Celsius or 32 Fahrenheit)273{\text{ }}K{\text{ }}\left( {0^\circ Celsius{\text{ }}or{\text{ }}32^\circ {\text{ }}Fahrenheit} \right) and therefore the standard atmosphere is 1 atm1{\text{ }}atm pressure.
Today you're getting to prove experimentally that the quantity of 1mole1mole of a gas at degree Centigrade & pressure (STP)\left( {STP} \right) occupies a volume of 224 liters{\text{22}} \cdot 4{\text{ }}liters or 22,400 milliliters22,400{\text{ }}milliliters . The numerical values that are used for STPSTP are one atmosphere (1 atm)\left( {1{\text{ }}atm} \right) and 00 degrees Celsius (0C) or 273 Kelvin (273K)\left( {0^\circ C} \right){\text{ }}or{\text{ }}273{\text{ }}Kelvin{\text{ }}\left( {273K} \right)

Note:
Degree Centigrade and Pressure. Degree Centigrade is adequate to 0C0^\circ C , which is 273.15 K273.15{\text{ }}K . Standard atmosphere is 1atm1atm , 101.3kPa or 760 mmHg or torr101.3kPa{\text{ }}or{\text{ }}760{\text{ }}mmHg{\text{ }}or{\text{ }}torr . STPSTP is the "standard" condition often used for measuring gas density and volume.