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Chapter 9: Problem 1227

\- decrease the volume of gas by \(\supset \%\) at constant temperature the pressure should be (A) Increased by \(5.26 \%\) (B) Decreased by \(5.26 \%\) (C) Decreased by \(11 \%\) (D) Increased by \(11 \%\)

Short Answer

Expert verified
The pressure of the gas will increase by approximately \(11 \%\) when the volume is decreased by \(10 \%\) at constant temperature. The correct answer is (D) Increased by \(11 \%\).

Step by step solution

01

Recall Boyle's Law

Boyle's Law states that Pressure(P) is inversely proportional to Volume(V) at constant temperature(T). Mathematically, it can be represented as: \[P \propto \frac{1}{V}\]
02

Express the relationship as an equation

To solve the problem, eliminate the proportionality and add a constant (k). So, we have: \[ P = \frac{k}{V}\] where k is the proportionality constant.
03

Determine the initial and final volumes

Since the volume is decreased by 10%, the initial volume (V1) can be represented as: \[V1 = V\] And the final volume (V2) can be represented as: \[V2 = 0.9V\] where V represents the original volume of the gas.
04

Apply Boyle's Law for initial and final states

By applying Boyle's Law to both the initial and final states, we have following equations: For initial state, \[P1 = \frac{k}{V}\] For final state, \[P2 = \frac{k}{0.9V}\]
05

Eliminate the constant k

We can eliminate the constant k by dividing the equation of the final state by the equation of the initial state: \[\frac{P2}{P1} = \frac{\frac{k}{0.9V}}{\frac{k}{V}}\]
06

Solve for the ratio of final to initial pressure

Simplify the equation to find the ratio of the final pressure (P2) to the initial pressure (P1): \[\frac{P2}{P1} = \frac{V}{0.9V}\] Simplifying further, we get: \[\frac{P2}{P1} = \frac{1}{0.9}\]
07

Find the percentage increase in pressure

Now find the percentage increase in pressure by dividing the final pressure by the initial pressure and then subtracting 1: Percentage increase = \(\frac{P2 - P1}{P1} \times 100\%) Use the ratio P2/P1 to get: Percentage increase = \(\frac{\frac{1}{0.9} - 1}{1} \times 100\%\) Simplify further to obtain the percentage increase in pressure: Percentage increase = \(\frac{0.1}{0.9} \times 100\% \approx 11.11\%\) So, the pressure has increased by approximately 11.11%. The correct answer is (D) Increased by \(11 \%\).

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At what temperature pressure remaining constant will the \(\mathrm{rms}\) speed of a gas molecules increase by \(10 \%\) of the \(\mathrm{rms}\) speed at NTP? (A) \(57.3 \mathrm{~K}\) (B) \(57.3^{\circ} \mathrm{C}\) (C) \(557.3 \mathrm{~K}\) (D) \(-57.3^{\circ} \mathrm{C}\)
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