Question

1. a neon sign is made from a glass tube 5.5 m long and with an inner diameter of 2.5 cm. if the neon inside is at a pressure of 1.78 torr at 35°c, how many grams of neon are in the sign?

Explanation:

Step1: Convert units to SI units

First, convert the length of the tube to centimeters: \( 5.5\ m = 550\ cm \). The inner diameter is \( 2.5\ cm \), so the radius \( r=\frac{2.5}{2}=1.25\ cm \). The volume of a cylinder is \( V = \pi r^{2}h \), where \( h \) is the height (length) of the cylinder. So \( V=\pi\times(1.25)^{2}\times550 \approx 2699.09\ cm^{3} \). Convert volume to liters: \( 2699.09\ cm^{3}=2.69909\ L \) (since \( 1\ L = 1000\ cm^{3} \)).

Convert pressure from torr to atm: \( 1\ atm = 760\ torr \), so \( P=\frac{1.78}{760}\ atm\approx0.002342\ atm \).

Convert temperature from Celsius to Kelvin: \( T = 35 + 273.15 = 308.15\ K \).

Step2: Use the ideal gas law \( PV = nRT \) to find moles

The ideal gas law is \( PV = nRT \), where \( R = 0.0821\ L\cdot atm/(mol\cdot K) \). Solve for \( n \): \( n=\frac{PV}{RT} \).

Substitute the values: \( n=\frac{0.002342\ atm\times2.69909\ L}{0.0821\ L\cdot atm/(mol\cdot K)\times308.15\ K} \).

Calculate the numerator: \( 0.002342\times2.69909\approx0.006322 \).

Calculate the denominator: \( 0.0821\times308.15\approx25.299 \).

So \( n=\frac{0.006322}{25.299}\approx0.00025\ mol \).

Step3: Calculate mass from moles

The molar mass of neon (Ne) is \( 20.18\ g/mol \). Mass \( m = n\times M \), so \( m = 0.00025\ mol\times20.18\ g/mol\approx0.005045\ g \). Wait, let's recalculate the moles more accurately.

Wait, let's redo the mole calculation:

\( n=\frac{PV}{RT}=\frac{(\frac{1.78}{760})\times2.69909}{0.0821\times308.15} \)

First, \( \frac{1.78}{760}=0.0023421 \)

\( 0.0023421\times2.69909 = 0.006322 \)

\( 0.0821\times308.15 = 25.299 \)

\( n=\frac{0.006322}{25.299}\approx0.00025\ mol \)? Wait, no, that seems too low. Wait, maybe I made a mistake in volume calculation.

Wait, length is 5.5 m = 550 cm, diameter 2.5 cm, radius 1.25 cm. Volume of cylinder: \( V=\pi r^{2}h=\pi\times(1.25)^2\times550 \). Let's calculate that again: \( 1.25^2 = 1.5625 \), \( 1.5625\times550 = 859.375 \), \( 859.375\times\pi\approx2699.11\ cm^3 = 2.69911\ L \). That's correct.

Pressure: 1.78 torr. 1 torr = 1/760 atm, so 1.78/760 ≈ 0.002342 atm. Correct.

Temperature: 35 + 273.15 = 308.15 K. Correct.

Ideal gas law: \( n = PV/(RT) \). So \( P = 0.002342\ atm \), \( V = 2.69911\ L \), \( R = 0.0821\ L·atm/(mol·K) \), \( T = 308.15\ K \).

So \( n = (0.002342 * 2.69911) / (0.0821 * 308.15) \).

Calculate numerator: 0.002342 * 2.69911 ≈ 0.006322.

Denominator: 0.0821 * 308.15 ≈ 25.299.

So \( n ≈ 0.006322 / 25.299 ≈ 0.00025\ mol \)? Wait, that can't be right. Wait, maybe I messed up the pressure conversion. Wait, 1 torr is 1 mm Hg, and ideal gas constant R can also be used with torr: R = 62.3637 L·torr/(mol·K). Let's try that.

So \( R = 62.3637\ L·torr/(mol·K) \), \( P = 1.78\ torr \), \( V = 2.69911\ L \), \( T = 308.15\ K \).

Then \( n = PV/(RT) = (1.78 * 2.69911) / (62.3637 * 308.15) \).

Calculate numerator: 1.78 * 2.69911 ≈ 4.804.

Denominator: 62.3637 * 308.15 ≈ 19217.

So \( n ≈ 4.804 / 19217 ≈ 0.00025\ mol \). Wait, that's the same. Then mass is m = n * M, M for Ne is 20.18 g/mol. So 0.00025 mol * 20.18 g/mol ≈ 0.005045 g. But that seems very small. Wait, maybe the pressure is 1.78 torr, which is very low, so the amount is small. Alternatively, maybe I made a mistake in volume. Wait, 5.5 m is 550 cm, diameter 2.5 cm, so radius 1.25 cm. Volume: πr²h = π*(1.25)^2*550. Let's compute 1.25^2 = 1.5625, 1.5625*550 = 859.375, 859.375*π ≈ 2699 cm³ = 2.699 L. That's correct.

Wait, let's check the ideal gas law again. The formula is correct. So with P in torr, V in L, R = 62.36 L·torr/(mol·K), T in K. So n = (P*V)/(R*T) = (1.78 * 2.699) / (62.36 * 308.15). Let's compute that:

1.78 * 2.699 ≈ 4.804

62.36 * 308.15 ≈ 62.36*300=18708, 62.36*8.15≈508.234, total≈18708+508.234=19216.234

So n≈4.804/19216.234≈0.00025 mol. Then mass is 0.00025 * 20.18 ≈ 0.0050 g. So approximately 0.005 grams. But let's check with more precise calculations.

Alternatively, maybe the pressure is 1.78 torr, which is 1.78/760 atm ≈ 0.002342 atm. V=2.699 L, T=308.15 K, R=0.0821 L·atm/(mol·K). So n=(0.002342 * 2.699)/(0.0821 * 308.15) = (0.006322)/(25.299) ≈ 0.00025 mol. So mass is 0.00025 * 20.18 ≈ 0.0050 g. So the answer is approximately 0.005 grams. Wait, but maybe I made a mistake in volume. Wait, 5.5 meters is 550 centimeters, diameter 2.5 cm, so radius 1.25 cm. Volume of cylinder: πr²h = 3.1416 * 1.25² * 550. 1.25²=1.5625, 1.5625*550=859.375, 859.375*3.1416≈2699.11 cm³=2.69911 L. That's correct.

So the steps are:

1. Calculate volume of the tube (cylinder) using \( V = \pi r^2 h \).

2. Convert pressure from torr to atm (or use R in torr units).

3. Convert temperature to Kelvin.

4. Use ideal gas law \( PV = nRT \) to find moles (n).

5. Calculate mass using \( m = n \times M \), where M is molar mass of Ne (20.18 g/mol).

Answer:

Approximately \(\boxed{0.005}\) grams (or more precisely, around 0.0050 grams).