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Why you use kelivin in calculation instead of celsius and Fahrenheit?
Wiki User
∙ 10y ago
Kelvin has the advantage that it is an absolute temperature scale - it starts from absolute zero. This simplifies several calculations; for example, in an ideal gas, at constant pressure, the volume of the gas is proportional to the absolute temperature. Similarly, calculations related to heat machines are simpler if an absolute temperature scale is used.
Wiki User
∙ 10y ago
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Why is a temperature difference the same in celsius and kelvin?
Because degree Kelivin (K) = degree Celsius (C) + 273 So, K1 = C1 + 273 K2 = C2 + 273 K1 - K2 = C1 - C2; same difference
What pressure is argon at room temperature?
The given information is not sufficient to answer this question. You can use the Ideal Gas Law to find out though, expressed mathematically as: PV=nRT Where: P=Pressure (in mmHg) V=Volume (in Liters) n=number of moles of gas R=62.36367 L·mmHg·K−1·mol−1 (Ignore the jargon at the end just know that solving the equation for P will give an answer in the unit mmHg,) T= Temperature (in Kelvin) (room temperature in Kelivin is 293 K) You would already need to know V and n to begin with in order to be able to do this equation, however for the sake of example (exactly) one liter and .0094 mole of Argon would be: P(1)=(.0094)(62.36367)(293) P= 170 mmHg One mole or 39.948 grams of Argon would be at a pressure of P(1)=(1)(62.36367)(293) P= 18300 mmHg which is 24 times the pressure of Earth's atmosphere.
What pressure is krypton at room temperature?
The given information is not sufficient to answer this question. You can use the Ideal Gas Law to find out though, expressed mathematically as: PV=nRT Where: P=Pressure (in mmHg) V=Volume (in Liters) n=number of moles of gas R=62.36367 L·mmHg·K−1·mol−1 (Ignore the jargon at the end just know that solving the equation for P will give an answer in the unit mmHg,) T= Temperature (in Kelvin) (room temperature in Kelivin is 293 K) You would already need to know V and n to begin with in order to be able to do this equation, however for the sake of example (exactly) one liter and .0094 mole of Argon would be: P(1)=(.0094)(62.36367)(293) P= 170 mmHg One mole or 39.948 grams of Argon would be at a pressure of P(1)=(1)(62.36367)(293) P= 18300 mmHg which is 24 times the pressure of Earth's atmosphere (760 mmHg).
