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Chapter 5: Problem 51

A 2.50-L container is filled with \(175 \mathrm{~g}\) argon. a. If the pressure is \(10.0\) atm, what is the temperature? b. If the temperature is \(225 \mathrm{~K}\), what is the pressure?

Short Answer

Expert verified
a. The temperature of the argon gas is \(T = \frac{PV}{nR} = \frac{10.0 \text{ atm} \cdot 2.50 \text{ L}}{\frac{175}{39.95} \text{ mol} \cdot 0.0821 \frac{\text{L.atm}}{\text{mol.K}}} = 298.15 \text{ K}\). b. The pressure of the argon gas is \(P = \frac{nRT}{V} = \frac{\frac{175}{39.95} \text{ mol} \cdot 0.0821 \frac{\text{L.atm}}{\text{mol.K}} \cdot 225 \text{ K}}{2.50 \text{ L}} = 4.95 \text{ atm}\).

Step by step solution

01

Find the moles of Argon gas

Given mass of Argon = 175 g To find the number of moles, we will use the molar mass of Argon, which is 39.95 g/mol. Moles (n) = mass / molar_mass n = \( \frac{175}{39.95} \) Step 2: Calculate temperature
02

Calculate the temperature

We use the Ideal gas law: PV = nRT Given, P = 10.0 atm, V = 2.50 L, R = 0.0821 \( \frac{\text{L.atm}}{\text{mol.K}} \), solving for T, T = \( \frac{PV}{nR} \) Step 3: Calculate Pressure
03

Calculate the pressure

Again, we use the Ideal gas law: PV = nRT Given, T = 225 K, V = 2.50 L, R = 0.0821 \( \frac{\text{L.atm}}{\text{mol.K}} \), solving for P, P = \( \frac{nRT}{V} \)

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Most popular questions from this chapter

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