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Chapter 14: Problem 28

A chamber with a fixed volume of \(1.0 \mathrm{m}^{3}\) contains a monatomic gas at \(3.00 \times 10^{2} \mathrm{K} .\) The chamber is heated to a temperature of \(4.00 \times 10^{2} \mathrm{K}\). This operation requires \(10.0 \mathrm{J}\) of heat. (Assume all the energy is transferred to the gas.) How many gas molecules are in the chamber?

Short Answer

Expert verified
Answer: There are approximately \(1.087 \times 10^{21}\) gas molecules in the chamber.

Step by step solution

01

Write down the given values

We have the following given quantities: Initial Temperature (T1) = \(3.00 \times 10^{2} \mathrm{K}\) Final Temperature (T2) = \(4.00 \times 10^{2} \mathrm{K}\) Heat (Q) = \(10.0 \mathrm{J}\) Volume (V) = \(1.0 \mathrm{m}^3 = 1000 \mathrm{L}\)
02

Calculate the heat capacity for a monatomic gas

The heat capacity for a monatomic gas at constant volume (Cv) can be calculated using the following equation: \(C_v = \frac{3}{2}R\), where R represents the ideal gas constant which is \(8.314 \mathrm{J \cdot mol^{-1}\cdot K^{-1}}\).
03

Calculate the number of moles of the gas

Now, we can use the heat capacity equation to find the number of moles of the gas. The equation for the heat capacity is: \(Q = n C_v \Delta T\), where n represents the number of moles, and \(\Delta T\) is the change in temperature. We can rearrange this equation to solve for n: \(n = \frac{Q}{C_v \Delta T}\). Plugging in the known values: \(n = \frac{10}{\frac{3}{2} \cdot 8.314 \cdot (4.00 \times 10^{2} - 3.00 \times 10^{2})}\) Calculating the number of moles: \(n \approx 1.805 \times 10^{-3} \mathrm{mol}\).
04

Calculate the number of gas molecules

Finally, we will use Avogadro's number to find the number of gas molecules in the chamber. Avogadro's number (N) is \(6.022 \times 10^{23}\) molecules per mole. Number of gas molecules = n × N Number of gas molecules = \(1.805 \times 10^{-3} \mathrm{mol} \times 6.022 \times 10^{23} \mathrm{molecules/mol}\) Number of gas molecules \(\approx 1.087 \times 10^{21}\) There are approximately \(1.087 \times 10^{21}\) gas molecules in the chamber.

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Most popular questions from this chapter

A \(0.400-\mathrm{kg}\) aluminum teakettle contains \(2.00 \mathrm{kg}\) of water at \(15.0^{\circ} \mathrm{C} .\) How much heat is required to raise the temperature of the water (and kettle) to \(100.0^{\circ} \mathrm{C} ?\)
In a physics lab, a student accidentally drops a \(25.0-\mathrm{g}\) brass washer into an open dewar of liquid nitrogen at 77.2 K. How much liquid nitrogen boils away as the washer cools from \(293 \mathrm{K}\) to $77.2 \mathrm{K} ?\( The latent heat of vaporization for nitrogen is \)199.1 \mathrm{kJ} / \mathrm{kg}$.
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