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Each of the small metal canisters shown in the photograph contains 0.36 moles of carbon dioxide - Leaving Cert Chemistry - Question f - 2020

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Each of the small metal canisters shown in the photograph contains 0.36 moles of carbon dioxide. The contents of one canister were released into a flat bicycle tube.... show full transcript

Worked Solution & Example Answer:Each of the small metal canisters shown in the photograph contains 0.36 moles of carbon dioxide - Leaving Cert Chemistry - Question f - 2020

Step 1

Using the Ideal Gas Law: $PV = nRT$

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Answer

To calculate the pressure of the gas inside the tube, we start by using the Ideal Gas Law, given by:

PV=nRTPV = nRT

where:

  • PP is the pressure,
  • VV is the volume,
  • nn is the number of moles of gas,
  • RR is the universal gas constant (approximately 8.31 J/(mol·K)), and
  • TT is the temperature in Kelvin.

Step 2

Substituting Values

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Answer

We have:

  • n=0.36n = 0.36 moles of CO2CO_2,
  • V=1.0×103m3V = 1.0 \times 10^{-3} \, m^3,
  • R=8.31J/(molK)R = 8.31 \, J/(mol \, K),
  • T=293KT = 293 \, K.

Substituting these values into the Ideal Gas Law:

P×(1.0×103)=0.36×8.31×293P \times (1.0 \times 10^{-3}) = 0.36 \times 8.31 \times 293

Step 3

Calculating Pressure

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Answer

Now we can calculate the right-hand side:

0.36×8.31×293=87654.81Pa0.36 \times 8.31 \times 293 = 87654.81 \, Pa

Thus, we rearrange the formula to solve for PP:

P=87654.811.0×103=8.765481×107PaP = \frac{87654.81}{1.0 \times 10^{-3}} = 8.765481 \times 10^7 \, Pa

To express this to two significant figures, the pressure of the gas inside the tube is:

P8.8×105PaP \approx 8.8 \times 10^5 \, Pa

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