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Chapter 17: Problem 21

(a) If 2.0 mol of an ideal gas are initially at temperature \(250 \mathrm{K}\) and pressure 1.5 atm, what's the gas volume? (b) The pressure is now increased to 4.0 atm, and the gas volume drops to half its initial value. What's the new temperature?

Short Answer

Expert verified
The initial volume of the gas is approximately 27.36 L. After the pressure increase and volume decrease, the new temperature of the gas is approximately 498 K.

Step by step solution

01

Initial Volume Calculation

The ideal gas law can be rearranged to solve for volume as \(V = \frac{nRT}{P}\). Let's use moles (n) as 2.0, Temp (T) as 250K, pressure (P) as 1.5 atm, and R (the ideal gas constant) as 0.08206 L路atm/K路mol. The volume (V) can be calculated as \(V = \frac{(2.0 \, (\text{mol}) \times 0.08206 \, (\text{L路atm/K路mol}) \times 250 \, (\text{K})}{1.5 \, (\text{atm})}\).
02

New Temperature Calculation

Applying the ideal gas law again, we can find the new temperature (T') using \(T' = \frac{P'V'}{nR}\), where P' is the new pressure, V' is the new volume, and n and R are the same as before. The new pressure (P') is 4.0 atm and the new volume (V') is half of the initial volume. So, \(T' = \frac{4.0 \, (\text{atm}) \times \frac{V}{2}}{2.0 \, (\text{mol}) \times 0.08206 \, (\text{L路atm/K路mol})}\).
03

Calculating the new temperature using previous step

Replace V in the formula above with the initial volume calculated in step 1.

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