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Chapter 11: Problem 47

According to the ideal gas law, what would happen to the pressure of a gas if you doubled the amount of gas in a container while also tripling the Kelvin temperature of the gas? Explain.

Short Answer

Expert verified
The pressure of the gas will increase by a factor of 6 if the amount of gas is doubled and the Kelvin temperature is tripled, assuming a constant volume. This is based on the ideal gas law equation \(PV = nRT\), and by applying the given changes, we find that \(P_{new} = 6P\).

Step by step solution

01

The ideal gas law is given by the equation: \(PV = nRT\), where P is pressure, V is volume, n is the amount of gas, R is the ideal gas constant, and T is the temperature in Kelvin. #Step 2: Establish the given changes in the gas and temperature#

According to the exercise, we have two changes: 1. The amount of gas (n) is doubled, so the new amount of gas is \(2n\). 2. The Kelvin temperature (T) is tripled, so the new temperature is \(3T\). #Step 3: Apply the changes to the ideal gas law#
02

To find the new pressure after applying the given changes, we can substitute the new values of n and T into the ideal gas law: \(P_{new}V = (2n)R(3T)\). #Step 4: Introduce a constant volume#

Since we're not given any information about how volume (V) is changing, we can assume it remains constant. This means the original and new pressure can be compared at the same volume. We can therefore divide the new equation by the original equation: \(\frac{P_{new}V}{PV} = \frac{(2n)(3T)(R)}{nTR}\). #Step 5: Simplify and solve for the new pressure#
03

Simplify the equation by canceling out common terms: \(\frac{P_{new}}{P} = \frac{6nT}{nT}\). Now, the n and T terms cancel out: \(\frac{P_{new}}{P} = 6\) Finally, solve for the new pressure: \(P_{new} = 6P\) #Step 6: Interpret the result#

The result shows that the pressure of the gas (P) will increase by a factor of 6 if the amount of gas (n) is doubled and the Kelvin temperature (T) is tripled while keeping volume constant.

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Most popular questions from this chapter

Consider two identical \(1-\mathrm{L}\) containers, both at room temperature \((300 \mathrm{~K})\). One of them contains 1.0 \(\mathrm{g}\) of helium gas, and the other contains \(1.0 \mathrm{~g}\) of hydrogen gas. Is the pressure higher in the helium container, higher in the hydrogen container, or the same in the two containers?

(a) If the temperature of a gas is doubled while the pressure is kept constant, the volume of the gas (b) If the pressure of a gas is halved while the temperature is kept constant, the volume of the gas

How do you convert from \({ }^{\circ} \mathrm{C}\) to \(\mathrm{K}\) ? Convert room temperature \(\left(22.0^{\circ} \mathrm{C}\right)\) to Kelvin.

Rewrite the ideal gas law solving for \(T\). Also show how all units cancel to leave you with just units of temperature.

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 \(4 \pi r^{2}\) ).
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