Question

mr. pauller chemistry 2 period name
ideal gas law problems
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ideal gas law
pv=nrt
r = 0.0821 l·atm/(mol·k)
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1. what is the volume of 4.5 moles of argon at a pressure of 2.8 atmospheres and temperature of 266 kelvin?
2. what is the pressure of 3.0 mol of nitrogen with a volume of 19.3 l at a temperature of 45°c?
3. how many moles of chlorine would occupy a 40 l barrel at 353 k when the pressure is 1.8 atm?
4. what would the temperature be of 1.0 mol of hydrogen in a 1.0 l container at 1.0 atm?

Explanation:

Problem 1 (Volume of Argon):

Step1: Identify known values

We know the ideal gas law is \( PV = nRT \). For Argon: \( n = 4.5 \, \text{mol} \), \( P = 2.8 \, \text{atm} \), \( T = 266 \, \text{K} \), \( R = 0.0821 \, \frac{\text{L·atm}}{\text{mol·K}} \). We need to find \( V \). Rearrange the formula to \( V=\frac{nRT}{P} \).

Step2: Substitute values

Substitute \( n = 4.5 \), \( R = 0.0821 \), \( T = 266 \), \( P = 2.8 \) into the formula:
\( V=\frac{4.5 \times 0.0821 \times 266}{2.8} \)
First, calculate the numerator: \( 4.5\times0.0821 = 0.36945 \); \( 0.36945\times266 = 98.2737 \)
Then divide by \( P \): \( V=\frac{98.2737}{2.8}\approx35.1 \, \text{L} \)

Problem 2 (Pressure of Nitrogen):

Step1: Identify known values

For Nitrogen: \( n = 3.0 \, \text{mol} \), \( V = 19.2 \, \text{L} \), \( T = 45^\circ\text{C} = 45 + 273.15 = 318.15 \, \text{K} \), \( R = 0.0821 \, \frac{\text{L·atm}}{\text{mol·K}} \). Rearrange \( PV = nRT \) to \( P=\frac{nRT}{V} \).

Step2: Substitute values

Substitute \( n = 3.0 \), \( R = 0.0821 \), \( T = 318.15 \), \( V = 19.2 \) into the formula:
\( P=\frac{3.0\times0.0821\times318.15}{19.2} \)
Numerator: \( 3.0\times0.0821 = 0.2463 \); \( 0.2463\times318.15\approx78.36 \)
Then divide by \( V \): \( P=\frac{78.36}{19.2}\approx4.08 \, \text{atm} \)

Problem 3 (Moles of Chlorine):

Step1: Identify known values

For Chlorine: \( V = 40 \, \text{L} \), \( T = 333 \, \text{K} \), \( P = 1.8 \, \text{atm} \), \( R = 0.0821 \). Rearrange \( PV = nRT \) to \( n=\frac{PV}{RT} \).

Step2: Substitute values

Substitute \( P = 1.8 \), \( V = 40 \), \( R = 0.0821 \), \( T = 333 \) into the formula:
\( n=\frac{1.8\times40\times0.0821}{333} \)
First, calculate the numerator: \( 1.8\times40 = 72 \); \( 72\times0.0821 = 5.9112 \)
Then divide by \( T \): \( n=\frac{5.9112}{333}\approx0.0178 \, \text{mol} \) (or more precisely, let's recalculate: \( 1.8\times40 = 72 \); \( 72\times0.0821 = 5.9112 \); \( 5.9112\div333\approx0.01775 \approx 0.018 \, \text{mol} \))

Problem 4 (Temperature of Hydrogen):

Step1: Identify known values

For Hydrogen: \( n = 1.0 \, \text{mol} \), \( V = 1.0 \, \text{L} \), \( P = 1.0 \, \text{atm} \), \( R = 0.0821 \). Rearrange \( PV = nRT \) to \( T=\frac{PV}{nR} \).

Step2: Substitute values

Substitute \( P = 1.0 \), \( V = 1.0 \), \( n = 1.0 \), \( R = 0.0821 \) into the formula:
\( T=\frac{1.0\times1.0\times1.0}{1.0\times0.0821} \) (Wait, no: \( P = 1.0 \, \text{atm} \), \( V = 1.0 \, \text{L} \), \( n = 1.0 \, \text{mol} \), so \( T=\frac{PV}{nR}=\frac{1.0\times1.0}{1.0\times0.0821}\approx12.18 \, \text{K} \)? Wait, no, that can't be right. Wait, no, \( PV = nRT \), so \( T=\frac{PV}{nR} \). So \( P = 1.0 \, \text{atm} \), \( V = 1.0 \, \text{L} \), \( n = 1.0 \, \text{mol} \), \( R = 0.0821 \). So \( T=\frac{1.0\times1.0}{1.0\times0.0821}\approx12.18 \, \text{K} \)? But that seems very low. Wait, maybe a typo? Wait, if \( V \) was 22.4 L, it would be 273 K, but here \( V = 1.0 \, \text{L} \), so the calculation is correct. So \( T=\frac{1.0\times1.0}{1.0\times0.0821}\approx12.2 \, \text{K} \).

Answer:

s:

1. Volume of Argon: \(\approx 35.1 \, \text{L}\)
2. Pressure of Nitrogen: \(\approx 4.08 \, \text{atm}\)
3. Moles of Chlorine: \(\approx 0.018 \, \text{mol}\)
4. Temperature of Hydrogen: \(\approx 12.2 \, \text{K}\)