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Chapter 11: Problem 67

The Kelvin temperature of a gas is doubled. At the same time, its volume is cut in half by compressing it with a piston. What happens to the pressure of the gas? (Hint: Set \(T_{f}=2 T_{i}\) and \(\left.V_{f}=0.5 V_{i}\right)\)

Short Answer

Expert verified
The pressure of the gas increases four times when the Kelvin temperature is doubled, and the volume is cut in half (\(P_f = 4P_i\)).

Step by step solution

01

Write the ideal gas law for the initial and final states

Write the ideal gas law for the initial state, letting \(P_i\) and \(V_i\) be the initial pressure and volume, respectively: \(P_iV_i = nRT_i\) Now, write the ideal gas law for the final state, letting \(P_f\) and \(V_f\) be the final pressure and volume, respectively: \(P_fV_f = nRT_f\)
02

Substitute the given temperature and volume relationships

We are given that \(T_f = 2T_i\) and \(V_f = 0.5V_i\). Substitute these relationships into the ideal gas law equation for the final state: \(P_f(0.5V_i) = nR(2T_i)\)
03

Solve for P_f

Now, our goal is to solve the equation for \(P_f\). First, divide both sides of the equation by \(0.5V_i\): \[ \frac{P_f(0.5V_i)}{0.5V_i} = \frac{nR(2T_i)}{0.5V_i} \] This will simplify to: \[ P_f = \frac{2nRT_i}{0.5V_i} \]
04

Relate initial pressure to initial temperature and volume

Using the initial state ideal gas equation, \(P_iV_i = nRT_i\), we can now relate \(P_i\) to \(T_i\) and \(V_i\). Divide both sides of the equation by \(V_i\): \[ \frac{P_iV_i}{V_i} = \frac{nRT_i}{V_i} \] This simplifies to: \[ P_i = \frac{nRT_i}{V_i} \]
05

Substitution and final answer

Now, substitute the expression for \(P_i\) from step 4, into the expression for \(P_f\) from step 3: \[ P_f = \frac{2nRT_i}{0.5V_i} \] \[ P_f = \frac{2\left(\dfrac{nRT_i}{V_i}\right)}{0.5} \] Since \(\frac{nRT_i}{V_i} = P_i\), we get: \[ P_f = \frac{2P_i}{0.5} \] Finally, this simplifies to: \[ P_f = 4P_i \] The pressure of the gas increases four times when the temperature is doubled, and the volume is halved.

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Most popular questions from this chapter

Explain why gases always occupy the entire container they are in, but solids and liquids do not.

Normal atmospheric pressure is \(1 \mathrm{~atm}=\) \(760 \mathrm{~mm} \mathrm{Hg} .\) However, it is also \(14.7 \mathrm{lbs} / \mathrm{in.}^{2}\) (14.7 pounds per square inch). Given that liquid mercury has a density of \(13.6 \mathrm{~g} / \mathrm{mL}\), explain (show a calculation) why \(1 \mathrm{~atm}\) also equals \(14.7 \mathrm{lbs} / \mathrm{in}^{2}\). Then, go back to the introduction of this chapter and calculate that the atmospheric force holding the Magdeburg spheres together is indeed close to the stated \(4.5\) tons (the surface area of a sphere is given by \(4 \pi r^{2}\) ).

A gas tank contains \(\mathrm{CO}_{2}\) at a pressure of \(6.80 \mathrm{~atm}\). What would the \(\mathrm{CO}_{2}\) pressure be if the container were (a) twice as large and (b) one-fourth as large?

What is the density in grams per liter of nitrogen gas at STP?

A chemical reaction produces \(100.0\) mL of hydrogen gas at \(745.5 \mathrm{~mm} \mathrm{Hg}\) and \(24.0^{\circ} \mathrm{C}\). What is the amount of \(\mathrm{H}_{2}\) produced in moles and in grams?
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