Question

1. what volume is needed to store...
2. what pressure will be exerted by 20.16g hydrogen gas in a 7.5l cylinder at 20°c?
3. a 50l cylinder is filled with argon gas to a pressure of 10130.0kpa at 30°c. how many moles of argon gas are in the cylinder?

Explanation:

Problem 2 (Volume Calculation, assuming ideal gas law and missing substance details, let's assume it's a gas with known moles, here we'll show the process with ideal gas law \( PV = nRT \))

Step1: Recall Ideal Gas Law

The ideal gas law is \( PV = nRT \), where \( P \) is pressure, \( V \) is volume, \( n \) is moles, \( R \) is gas constant (\( 8.314 \, \text{L·kPa/(mol·K)} \) or \( 0.0821 \, \text{L·atm/(mol·K)} \)), \( T \) is temperature in Kelvin. First, convert temperature to Kelvin: \( T = 20 + 273.15 = 293.15 \, \text{K} \). But we need to know \( n \) (moles) and \( P \) (pressure) or have more info. Wait, the problem says "store 0.050 moles" (assuming from partial text), let's proceed with \( n = 0.050 \, \text{mol} \), assume standard pressure \( P = 101.3 \, \text{kPa} \) (or 1 atm). Let's use \( R = 8.314 \, \text{L·kPa/(mol·K)} \). Rearrange for \( V \): \( V=\frac{nRT}{P} \)

Step2: Plug in Values

\( n = 0.050 \, \text{mol} \), \( R = 8.314 \, \text{L·kPa/(mol·K)} \), \( T = 293.15 \, \text{K} \), \( P = 101.3 \, \text{kPa} \)
\( V=\frac{0.050 \times 8.314 \times 293.15}{101.3} \)
Calculate numerator: \( 0.050 \times 8.314 = 0.4157 \); \( 0.4157 \times 293.15 \approx 121.8 \)
Divide by 101.3: \( V \approx \frac{121.8}{101.3} \approx 1.20 \, \text{L} \) (this is a sample calculation, adjust if actual \( n \) or \( P \) differs)

Step1: Find Moles of Hydrogen

Molar mass of \( \text{H}_2 \) is \( 2 \, \text{g/mol} \). Moles \( n=\frac{\text{mass}}{\text{molar mass}}=\frac{20.16 \, \text{g}}{2 \, \text{g/mol}} = 10.08 \, \text{mol} \)

Step2: Convert Temperature to Kelvin

\( T = 20 + 273.15 = 293.15 \, \text{K} \)

Step3: Apply Ideal Gas Law (\( PV = nRT \)), Solve for \( P \)

\( P=\frac{nRT}{V} \), \( R = 8.314 \, \text{L·kPa/(mol·K)} \), \( V = 7.5 \, \text{L} \)
\( P=\frac{10.08 \times 8.314 \times 293.15}{7.5} \)
Calculate numerator: \( 10.08 \times 8.314 \approx 83.81 \); \( 83.81 \times 293.15 \approx 24560 \)
Divide by 7.5: \( P \approx \frac{24560}{7.5} \approx 3275 \, \text{kPa} \) (or convert to atm if needed, but kPa is fine here)

Step1: Convert Temperature to Kelvin

\( T = 30 + 273.15 = 303.15 \, \text{K} \)

Step2: Apply Ideal Gas Law (\( PV = nRT \)), Solve for \( n \)

\( n=\frac{PV}{RT} \), \( P = 10130.0 \, \text{kPa} \), \( V = 50 \, \text{L} \), \( R = 8.314 \, \text{L·kPa/(mol·K)} \)
\( n=\frac{10130.0 \times 50}{8.314 \times 303.15} \)
Calculate numerator: \( 10130.0 \times 50 = 506500 \)
Denominator: \( 8.314 \times 303.15 \approx 2520 \)
\( n=\frac{506500}{2520} \approx 201 \, \text{mol} \) (approximate, more precise calculation: \( 8.314 \times 303.15 = 8.314\times300 + 8.314\times3.15 = 2494.2 + 26.19 = 2520.39 \); \( 506500\div2520.39\approx200.96\approx201 \, \text{mol} \))

Answer:

Approximately \( 1.20 \, \text{L} \) (depends on exact \( n \) and \( P \) values)

Problem 3 (Pressure Calculation for Hydrogen Gas)