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 empirical formula of the organic compound is =
Explanation:
At STP, 1 mole of gas occupies 22.4 L of volume.
Moles of gas at STP occupying 2.0 L = n
Moles of carbon in 0.08920 mol = 1 × 0.08920 mol = 0.08920 mol
Moles of gas at STP occupying 3.0 L = n'
Moles of hydrogen in 0.1339 moles of water vapor = 2 × 0.1339 mol = 0.2678 mol
Moles of gas at STP occupying 1.0 L = n''
Moles of sulfur in 0.04464 mol = 1 × 0.04464 mol = 0.04464 mol
Moles of carbon , hydrogen and sulfur constituent of that organic compound .
Moles of carbon in 0.08920 mol = 1 × 0.08920 mol = 0.08920 mol
Moles of hydrogen in 0.1339 moles of water vapor = 2 × 0.1339 mol = 0.2678 mol
Moles of sulfur in 0.04464 mol = 1 × 0.04464 mol = 0.04464 mol
For empirical; formula divide the least number of moles from all the moles of elements.
carbon =
Hydrogen =
Sulfur =
The empirical formula of the organic compound is =