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A helium balloon is inflated to a volume of 5.65 L and a pressure of 10.2 atm at a temperature of 25 °C - VCE - SSCE Chemistry - Question 16 - 2012 - Paper 1

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A helium balloon is inflated to a volume of 5.65 L and a pressure of 10.2 atm at a temperature of 25 °C. The amount of helium, in moles, in the balloon is A. 0.023 B... show full transcript

Worked Solution & Example Answer:A helium balloon is inflated to a volume of 5.65 L and a pressure of 10.2 atm at a temperature of 25 °C - VCE - SSCE Chemistry - Question 16 - 2012 - Paper 1

Step 1

Calculate the temperature in Kelvin

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Answer

To convert the temperature from Celsius to Kelvin, we use the formula: T(K)=T(°C)+273.15T(K) = T(°C) + 273.15 Substituting the given value: T=25+273.15=298.15KT = 25 + 273.15 = 298.15 K

Step 2

Use the Ideal Gas Law

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Answer

The Ideal Gas Law is given by: PV=nRTPV = nRT where:

  • P = pressure in atm = 10.2 atm
  • V = volume in L = 5.65 L
  • n = number of moles of gas
  • R = ideal gas constant = 0.0821 L·atm/(K·mol)
  • T = temperature in Kelvin = 298.15 K

Step 3

Rearrange the Ideal Gas Law to solve for n

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Answer

Rearranging the formula gives: n=PVRTn = \frac{PV}{RT}

Step 4

Substitute the values into the equation

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Substituting the known values into the equation: n=(10.2 atm)(5.65 L)(0.0821 L\cdotpatm/(K\cdotpmol))(298.15 K)n = \frac{(10.2 \text{ atm})(5.65 \text{ L})}{(0.0821 \text{ L·atm/(K·mol)})(298.15 \text{ K})}

Step 5

Calculate the number of moles

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Answer

Calculating it step-by-step:

  1. Calculate the numerator:
    • 10.2×5.65=57.6310.2 \times 5.65 = 57.63
  2. Calculate the denominator:
    • 0.0821×298.1524.4750.0821 \times 298.15 \approx 24.475
  3. Finally, the number of moles n is: n=57.6324.4752.35n = \frac{57.63}{24.475} \approx 2.35 This value rounds to 2.36 moles, which corresponds to option C.

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