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Chapter 9: Problem 1244

At constant temperature on increasing the pressure of a gas \(5 \%\) its volume will decrease by (A) \(5 \%\) (B) \(5.26 \%\) (C) \(4.76 \%\) (D) \(4.26 \%\)

Short Answer

Expert verified
The percentage decrease in volume when the pressure of a gas is increased by 5% at constant temperature is \(5\%\).

Step by step solution

01

Write down the Boyle's law formula

Boyle's law states that for a constant temperature process, the product of the initial pressure and initial volume is equal to the product of the final pressure and final volume. The formula for Boyle's law is: \[P_1V_1 = P_2V_2\]
02

Define the given values

Let's define the given values: Initial pressure, \(P_1 = 1\) The pressure is increased by \(5\%\), so the final pressure is \(1 + 0.05 = 1.05\), so \(P_2 = 1.05\)
03

Plug the given values in the formula

Now, let's plug in the given values into Boyle's law formula: \[P_1V_1 = P_2V_2\] \[(1)V_1=(1.05)V_2\] Now our goal is to find the percentage decrease in volume.
04

Solve for the final volume

To find the final volume, rearrange the equation by solving for \(V_2\): \[V_2 = \frac{V_1}{1.05}\]
05

Calculate the percentage decrease in volume

Now that we have an expression for the final volume, we can calculate the percentage decrease in volume, which is given by: Percentage decrease in volume = \[\frac{V_1 - V_2}{V_1} * 100\] Substitute the expression for \(V_2\) found in Step 4: Percentage decrease in volume = \[\frac{V_1 - \frac{V_1}{1.05}}{V_1} * 100\] Simplify the expression: Percentage decrease in volume = \[\frac{1 - \frac{1}{1.05}}{1} * 100\] Percentage decrease in volume = \[\frac{0.05}{1} * 100\] Percentage decrease in volume = \(5\%\) (Option A)

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Most popular questions from this chapter

To double the volume of a given mass at an ideal gas at $27^{\circ} \mathrm{C}$ keeping the pressure constant one must raise the temperature in degree centigrade (A) \(54^{\circ}\) (B) \(600^{\circ}\) (C) \(327^{\circ}\) (D) \(270^{\circ}\)

\- 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 \(11 \%\)

The relation between the gas pressure \(\mathrm{P}\) and average kinetic energy per unit volume \(E\) is (A) \(\mathrm{P}=(2 / 3) \mathrm{E}\) (B) \(P=(3 / 2) E\) (C) \(\mathrm{P}=\mathrm{E}\) (D) \(\mathrm{P}=(\mathrm{E} / 2)\)

\- If electron tube was sealed oft during manuracture at a pressure of $1 \times 10^{-7}\( torr at \)27^{\circ} \mathrm{C}\(. Its volume is \)100 \mathrm{~cm}^{3}$. The number of molecules that remain in the tube is (density of mercury is \(13.6 \mathrm{gcm}^{-3}\) ) \((1\) torr $=133 \mathrm{~Pa}$ ) (A) \(3.9 \times 10^{11}\) (B) \(3 \times 10^{16}\) (C) \(2 \times 10^{14}\) (D) \(7 \times 10^{11}\)

The rms. speed of the molecules of a gas in a vessel is $400 \mathrm{~ms}^{-1}$. If half of the gas leaks out, at constant temperature, the r.m.s speed of the remaining molecules will be (A) \(800 \mathrm{~ms}^{-1}\) (B) \(200 \mathrm{~ms}^{-1}\) (C) \(400 \sqrt{2} \mathrm{~ms}^{-1}\) (D) \(400 \mathrm{~ms}^{-1}\)
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