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A 100 mL sample of helium exerts a pressure of 1 atm at 10°C - VCE - SSCE Chemistry - Question 14 - 2005 - Paper 1

Question 14

A 100 mL sample of helium exerts a pressure of 1 atm at 10°C. The volume of the container is reduced to 50 mL and then the temperature is increased to 20°C. The pre... show full transcript

Worked Solution & Example Answer:A 100 mL sample of helium exerts a pressure of 1 atm at 10°C - VCE - SSCE Chemistry - Question 14 - 2005 - Paper 1

Step 1

Use the Ideal Gas Law

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Answer

We start with the Ideal Gas Law, which is given by the formula:

$PV = nRT$

In our case, we will apply the combined gas law since we're dealing with changes in pressure, volume, and temperature. This can be expressed as:

$\frac{P_1 V_1}{T_1} = \frac{P_2 V_2}{T_2}$

Where:

$P_1$ = initial pressure = 1 atm
$V_1$ = initial volume = 100 mL
$T_1$ = initial temperature in Kelvin = 10°C + 273.15 = 283.15 K
$P_2$ = final pressure (to be calculated)
$V_2$ = final volume = 50 mL
$T_2$ = final temperature in Kelvin = 20°C + 273.15 = 293.15 K

Step 2

Calculate Final Pressure

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Answer

Substituting the known values into the combined gas law, we have:

$\frac{(1 \: \text{atm}) (100 \: \text{mL})}{283.15 \: ext{K}} = \frac{P_2 (50 \: \text{mL})}{293.15 \: ext{K}}$

Now, cross-multiplying gives:

$P_2 = \frac{(1 \: \text{atm})(100 \: \text{mL})(293.15 \: ext{K})}{(283.15 \: ext{K})(50 \: \text{mL})}$

Calculating this yields:

$P_2 = \frac{100 \times 293.15}{50 \times 283.15} \approx 2.06 \: \text{atm}$

Step 3

Select Closest Answer

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Answer

The calculated final pressure is approximately 2.06 atm. The option that is closest to this value from the choices provided is:

C. 2

A 100 mL sample of helium exerts a pressure of 1 atm at 10°C - VCE - SSCE Chemistry - Question 14 - 2005 - Paper 1

Worked Solution & Example Answer:A 100 mL sample of helium exerts a pressure of 1 atm at 10°C - VCE - SSCE Chemistry - Question 14 - 2005 - Paper 1

Use the Ideal Gas Law

Calculate Final Pressure

Select Closest Answer

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