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Chapter 10: Problem 62

What happens to the pressure of a gas under the following conditions? a. The absolute temperature is halved and the volume doubles. b. Both the absolute temperature and the volume double. c. The absolute temperature increases by \(75 \%,\) and the volume decreases by \(50 \%\).

Short Answer

Expert verified
Answer: In Scenario (a), the pressure is reduced to 25% of its initial value. In Scenario (b), the pressure remains unchanged at 100% of its initial value. In Scenario (c), the pressure becomes 350% of its initial value.

Step by step solution

01

Scenario (a)

Given that the absolute temperature is halved, \(T_{final} = \frac{1}{2} T_{initial}\). Also, the volume doubles, so \(V_{final} = 2V_{initial}\). Since \(n\) and \(R\) remain constant, we can write: \(P_{initial}V_{initial}=nR T_{initial} \Rightarrow P_{final} (2V_{initial})=nR\left(\frac{1}{2}T_{initial}\right)\). Divide both sides by \(nR\) and multiply both sides by \(\frac{1}{2}\) to get: \(P_{final}=\frac{V_{initial}}{T_{initial}}\left(\frac{1}{2} \times \frac{1}{2} \right)\). Hence, the pressure becomes \(\frac{1}{4}\) of its initial value or it is reduced to \(25 \%\).
02

Scenario (b)

Given that both the absolute temperature and the volume double, \(T_{final} = 2T_{initial}\) and \(V_{final} = 2V_{initial}\). Since \(n\) and \(R\) remain constant, we can write: \(P_{initial}V_{initial}=nR T_{initial} \Rightarrow P_{final} (2V_{initial})=nR(2T_{initial})\). Divide both sides by \(nR \) and \(2V_{initial}\) to get: \(P_{final}= \frac{T_{initial}}{V_{initial}}\). Hence, the pressure remains unchanged; it stays at \(100\%\) of its initial value.
03

Scenario (c)

Given that the absolute temperature increases by \(75 \%\), \(T_{final} = 1.75T_{initial}\); and the volume decreases by \(50\%\), \(V_{final}=\frac{1}{2}V_{initial}\). Since \(n\) and \(R\) remain constant, we can write: \(P_{initial}V_{initial}=nR T_{initial} \Rightarrow P_{final} \left(\frac{1}{2}V_{initial}\right)=nR(1.75T_{initial})\). Divide both sides by \(nR\) and \(\frac{1}{2} V_{initial}\) to get: \(P_{final}= 3.5\frac{T_{initial}}{V_{initial}}\). Hence, the pressure becomes \(3.5\) times (or \(350 \%\)) of its initial value.

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