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Delicious77 [7]
3 years ago
10

A gas occupies a volume at 34.2 mL at a temperature of 15.0 C and a pressure of 800.0 torr. What will be the volume of this gas

at STP?
Chemistry
2 answers:
Nimfa-mama [501]3 years ago
5 0
The ideal gas law equation can be used to find the number of moles of the gas at given conditions
PV = nRT
where 
P - pressure - 800.0 torr x 133.3 Pa/torr = 106 640 Pa
V - volume - 34.2 x 10⁻³ m³
n - number of moles 
R - universal gas constant - 8.314 Jmol⁻¹K⁻¹
T - temperature in Kelvin - 15 °C + 273 = 288 K
substituting these values in the equation
106 640 Pa x 34.2 x 10⁻³ m³ = n x 8.314 Jmol⁻¹K⁻¹ x 288 K 
n = 1.52 mol

molar volume is where 1 mol of any gas occupies a volume of 22.4 L at STP
therefore if 1 mol occupies - 22.4 L
then 1.52 mol occupies - 22.4 L/mol x 1.52 mol = 34.05 L
volume of the gas at STP is 34.05 L
Grace [21]3 years ago
3 0
The answer is 34.1 mL.
Solution:
Assuming ideal behavior of gases, we can use the universal gas law equation
     P1V1/T1 = P2V2/T2
The terms with subscripts of one represent the given initial values while for terms with subscripts of two represent the standard states which is the final condition.
At STP, P2 is 760.0torr and T2 is 0°C or 273.15K. Substituting the values to the ideal gas expression, we can now calculate for the volume V2 of the gas at STP:
     (800.0torr * 34.2mL) / 288.15K = (760.0torr * V2) / 273.15K
     V2 = (800.0torr * 34.2mL * 273.15K) / (288.15K * 760.0torr)
     V2 = 34.1 mL
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Answer:

The equilbrium constant is 179.6

Explanation:

To solve this question we can use the equation:

ΔG = -RTlnK

<em>Where ΔG is Gibbs free energy = 12.86kJ/mol</em>

<em>R is gas constant = 8.314x10⁻³kJ/molK</em>

<em>T is absolute temperature = 298K</em>

<em>And K is equilibrium constant.</em>

Replacing:

12.86kJ/mol = -8.314x10⁻³kJ/molK*298K lnK

5.19 = lnK

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There is 96200 J or 96 kJ of heat released.

Explanation:

<u>Step 1: </u>Data given

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