QUESTION IMAGE
Question
if 0.173 moles of ar occupies 240. ml at 120.°c, what is the pressure (in torr)?
1 atm = 760 torr; r = 0.0821 l atm mol⁻¹ k⁻¹
○ 17.7 torr
○ 1.77×10⁴ torr
○ 5400 torr
○ 23.3 torr
Step1: Convert units
Convert volume from mL to L: \(240. \, \text{mL} = 0.240 \, \text{L}\)
Convert temperature from \(^\circ\text{C}\) to K: \(T = 120 + 273.15 = 393.15 \, \text{K}\)
Step2: Use Ideal Gas Law
Ideal Gas Law: \(PV = nRT\) → \(P = \frac{nRT}{V}\)
Substitute values: \(n = 0.173 \, \text{mol}\), \(R = 0.0821 \, \text{L atm mol}^{-1}\text{K}^{-1}\), \(T = 393.15 \, \text{K}\), \(V = 0.240 \, \text{L}\)
\(P = \frac{0.173 \times 0.0821 \times 393.15}{0.240}\)
Calculate: \(P \approx 23.3 \, \text{atm}\)
Step3: Convert atm to torr
\(1 \, \text{atm} = 760 \, \text{torr}\) → \(P = 23.3 \times 760 \approx 1.77 \times 10^4 \, \text{torr}\) Wait, no—wait, recalculate \(P\) in atm:
Wait, mistake earlier: \(n = 0.173\)? Wait, no, 0.173 moles? Wait, no, 0.173 moles is too much for 240 mL. Wait, maybe typo? Wait, no, let's recalculate:
Wait, \(n = 0.173\) mol? Wait, 0.173 mol in 0.24 L at 393 K:
\(P = \frac{0.173 \times 0.0821 \times 393.15}{0.240}\)
\(0.173 \times 0.0821 = 0.0142\); \(0.0142 \times 393.15 \approx 5.58\); \(5.58 / 0.240 \approx 23.25 \, \text{atm}\)
Then \(23.25 \, \text{atm} \times 760 \, \text{torr/atm} \approx 1.77 \times 10^4 \, \text{torr}\). Wait, but options: 1.77×10⁴ torr is an option. Wait, but let's check again. Wait, maybe \(n = 0.0173\)? No, the problem says 0.173 moles. Wait, maybe I misread. Wait, the problem: 0.173 moles of Ar? That seems high, but proceed.
Wait, no—wait, the correct calculation:
\(P = \frac{0.173 \times 0.0821 \times 393.15}{0.240}\)
\(0.173 \times 0.0821 = 0.0142033\)
\(0.0142033 \times 393.15 = 0.0142033 \times 393.15 \approx 5.584\)
\(5.584 / 0.240 \approx 23.27 \, \text{atm}\)
\(23.27 \times 760 = 17685.2 \approx 1.77 \times 10^4 \, \text{torr}\). So the correct option is \(1.77 \times 10^4 \, \text{torr}\).
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1.77×10⁴ torr (Option: 1.77×10⁴ torr)