The specific heats at constant pressure of some common gases are provided as a thirdorder polynomial: �;<<< = � + �� +
��C + ��E, with units J/(mol K). For methane (CH4) the coefficients are � = 19.89, � = 5.024 × 10NC, � = 1.269 × 10NO, � = −11.01 × 10NQ. At a production facility, the gas in a 50-liter tank is compressed to 3,600 psi (gage) during which time the temperature rises to 200°F. How much heat in J is given off as the gas cools to room temperature? Suppose the compressed gas is helium, how much heat would be given off in this case?
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Answer:
The volume will be 568.89 mL.
Explanation:
Boyle's law says that "The volume occupied by a given gaseous mass at constant temperature is inversely proportional to pressure"
Boyle's law is expressed mathematically as:
Pressure * Volume = constant
or P * V = k
Gay-Lussac's law indicates that when there is a constant volume, as the temperature increases, the pressure of the gas increases. And when the temperature is decreased, the pressure of the gas decreases. That is, the pressure of the gas is directly proportional to its temperature. Gay-Lussac's law can be expressed mathematically as follows:
Where P = pressure, T = temperature, K = Constant
Finally, Charles's law indicates that as the temperature increases, the volume of the gas increases and as the temperature decreases, the volume of the gas decreases. In summary, Charles's law is a law that says that when the amount of gas and pressure are kept constant, the quotient that exists between the volume and the temperature will always have the same value:
Combined law equation is the combination of three gas laws called Boyle's, Charlie's and Gay-Lusac's law:
Studying an initial state 1 and a final state 2, it is fulfilled:
In this case:
- P1= 960 mmHg
- V1= 550 mL
- T1= 200 C= 473 K (being 0 C=273 K)
- P2= 830 mmHg
- V2= ?
- T2= 150 C= 423 K
Replacing:
Solving:
V2= 568.9 mL
<u><em>The volume will be 568.89 mL.</em></u>