Question

In: Physics

A gas sample enclosed in a rigid metal container at room temperature (20.0∘C) has an absolute pressure p1. The container is immersed in hot water until it warms to 40.0∘C. What is the new absolute pressure p2?

Part A

A gas sample enclosed in a rigid metal container at room temperature (20.0∘C) has an absolute pressure p1. The container is immersed in hot water until it warms to 40.0∘C. What is the new absolute pressure p2?

Express your answer in terms of p1.

p2 =

1.0682p1

Part B

Nitrogen gas is introduced into a large deflated plastic bag. No gas is allowed to escape, but as more and more nitrogen is added, the bag inflates to accommodate it. The pressure of the gas within the bag remains at 1.00 atm and its temperature remains at room temperature(20.0∘C). How many moles n have been introduced into the bag by the time its volume reaches 22.4 L?

Express your answer in moles.

Solutions

Expert Solution

(A)

Write the equation \(P \propto T\) for the pressure \(P_{1}\) and the new absolute pressure for the pressure \(P_{2}\) using the equation \(P \propto T\) as follows:

\(\frac{P_{1}}{P_{2}}=\frac{T_{1}}{T_{2}}\)

Solve for \(P_{2}\)

\(P_{2}=\frac{P_{1} T_{2}}{T_{1}}\)

Now, substitute \(40.0^{\circ} \mathrm{C}\) for \(T_{2}\) and \(20^{\circ} \mathrm{C}\) for \(T_{1}\) in the equation \(P_{2}=\frac{P_{1} T_{2}}{T_{1}} .\)

\(P_{2}=\frac{P_{1}\left(40^{\circ} \mathrm{C}\right)}{\left(20^{\circ} \mathrm{C}\right)}\)

\(=\frac{P_{1}(40+273 \mathrm{~K})}{(20+273 \mathrm{~K})}\)

\(=1.0682 P_{1}\)

Part A

The new absolute pressure \(P_{2}\) in terms of \(P_{1}\) is equal to \(1.0682 P_{1}\).

At constant volume, the pressure is directly proportional to the absolute temperature of the gas.

(B)

Rearrange the equation \(P V=n R T\) as follows:

\(n=\frac{P V}{R T}\)

Substitute \(101.325 \mathrm{~Pa}\) for \(\mathrm{P}, 22.4 \mathrm{~m}^{3}\) for \(\mathrm{V}, 8.314 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K}\) for \(\mathrm{R},\) and \(20.0{ }^{\circ} \mathrm{C}\) for \(\mathrm{T}\) in the equation \(n=\frac{P V}{R T} .\)

\(n=\frac{(101.325 \mathrm{~Pa})\left(22.4 \mathrm{~m}^{3}\right)}{(8.314 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K})\left(20.0^{\circ} \mathrm{C}\right)}\)

\(=\frac{(101.325 \mathrm{~Pa})\left(22.4 \mathrm{~m}^{3}\right)}{(8.314 \mathrm{~J} / \mathrm{mol} \cdot \mathrm{K})(20.0+273 \mathrm{~K})}\)

\(=0.932\) moles

Part B The number of moles introduced into the bag is equal to 0.932 moles.

According to the ideal gas equation, the product of the gas's pressure and volume is proportional to the product of the number of moles and the absolute temperature of the gas.


Part A

The new absolute pressure \(P_{2}\) in terms of \(P_{1}\) is equal to \(1.0682 P_{1}\).

Part B

The number of moles introduced into the bag is equal to 0.932 moles.

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