Answer :
The pressure of 1.64×[tex]10^{-3[/tex] mol of argon in a 75.3 mL container at 30°C is approximately 178.63 torr.
To calculate the pressure (P) in torr, we can use the ideal gas law equation,[tex]\(PV = nRT\)[/tex], where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
Given:
[tex]\(V = 75.3 \, \text{mL} = 75.3 \times 10^{-3} \, \text{L}\)[/tex]
[tex]\(n = 1.64 \times 10^{-3} \, \text{mol}\)[/tex]
[tex]\(R = 0.0821 \, \text{L} \cdot \text{atm} \, \text{mol}^{-1} \cdot \text{K}^{-1}\)[/tex]
[tex]\(T = 30 + 273.15 \, \text{K}\)[/tex]
Substituting these values into the ideal gas law equation, we get:
[tex]\(P \cdot 75.3 \times 10^{-3} \, \text{L} = 1.64 \times 10^{-3} \, \text{mol} \cdot 0.0821 \, \text{L} \cdot \text{atm} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} \cdot (30 + 273.15) \, \text{K}\)[/tex]
Simplifying the equation, we find:
[tex]\(P = \frac{{1.64 \times 10^{-3} \, \text{mol} \cdot 0.0821 \, \text{L} \cdot \text{atm} \cdot \text{mol}^{-1} \cdot \text{K}^{-1} \cdot (30 + 273.15) \, \text{K}}}{{75.3 \times 10^{-3} \, \text{L}}}\)[/tex]
To evaluate the expression for the pressure (P), let's substitute the given values into the equation:
[tex]\[P = \frac{{1.64 \times 10^{-3} \, \text{{mol}} \cdot 0.0821 \, \text{{L}} \cdot \text{{atm}} \cdot \text{{mol}}^{-1} \cdot \text{{K}}^{-1} \cdot (30 + 273.15) \, \text{{K}}}}{{75.3 \times 10^{-3} \, \text{{L}}}}\][/tex]
First, let's calculate the value inside the parentheses:
[tex]\[30 + 273.15 = 303.15 \, \text{{K}}\][/tex]
Now, we can substitute this value into the equation:
[tex]\[P = \frac{{1.64 \times 10^{-3} \, \text{{mol}} \cdot 0.0821 \, \text{{L}} \cdot \text{{atm}} \cdot \text{{mol}}^{-1} \cdot \text{{K}}^{-1} \cdot 303.15 \, \text{{K}}}}{{75.3 \times 10^{-3} \, \text{{L}}}}\][/tex]
Simplifying further:
[tex]\[P = \frac{{0.01344804}}{{0.0753}}\][/tex]
Calculating the division:
[tex]\[P \approx 178.63 \, \text{{torr}}\][/tex]
Therefore, the pressure (P) is approximately 178.63 torr.
Learn more about argon here:
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