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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 脳 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 hiker is wearing wool clothing of \(0.50-\mathrm{cm}\) thickness to keep warm. Her skin temperature is \(35^{\circ} \mathrm{C}\) and the outside temperature is \(4.0^{\circ} \mathrm{C} .\) Her body surface area is \(1.2 \mathrm{m}^{2} .\) (a) If the thermal conductivity of wool is $0.040 \mathrm{W} /(\mathrm{m} \cdot \mathrm{K}),$ what is the rate of heat conduction through her clothing? (b) If the hiker is caught in a rainstorm, the thermal conductivity of the soaked wool increases to \(0.60 \mathrm{W} /(\mathrm{m} \cdot \mathrm{K})\) (that of water). Now what is the rate of heat conduction?
Convert \(1.00 \mathrm{kJ}\) to kilowatt-hours \((\mathrm{kWh})\).

How much internal energy is generated when a \(20.0-\mathrm{g}\) lead bullet, traveling at \(7.00 \times 10^{2} \mathrm{m} / \mathrm{s},\) comes to a stop as it strikes a metal plate?
A person of surface area \(1.80 \mathrm{m}^{2}\) is lying out in the sunlight to get a tan. If the intensity of the incident sunlight is $7.00 \times 10^{2} \mathrm{W} / \mathrm{m}^{2},$ at what rate must heat be lost by the person in order to maintain a constant body temperature? (Assume the effective area of skin exposed to the Sun is \(42 \%\) of the total surface area, \(57 \%\) of the incident radiation is absorbed, and that internal metabolic processes contribute another \(90 \mathrm{W}\) for an inactive person.)

Imagine a person standing naked in a room at \(23.0^{\circ} \mathrm{C}\) The walls are well insulated, so they also are at \(23.0^{\circ} \mathrm{C}\) The person's surface area is \(2.20 \mathrm{m}^{2}\) and his basal metabolic rate is 2167 kcal/day. His emissivity is \(0.97 .\) (a) If the person's skin temperature were \(37.0^{\circ} \mathrm{C}\) (the same as the internal body temperature), at what net rate would heat be lost through radiation? (Ignore losses by conduction and convection.) (b) Clearly the heat loss in (a) is not sustainable-but skin temperature is less than internal body temperature. Calculate the skin temperature such that the net heat loss due to radiation is equal to the basal metabolic rate. (c) Does wearing clothing slow the loss of heat by radiation, or does it only decrease losses by conduction and convection? Explain.
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