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An ideal gas is kept at constant pressure \(6.00 \times 10^{6} \mathrm{Pa},\) initial volume \(0.200 \mathrm{m}^{3}\) and temperature \(300.0 \mathrm{K}\). If the gas expands at constant pressure to a volume of \(0.600 \mathrm{m}^{3}\) find: (a) the work done by the gas; (b) the temperature of the gas at the new volume: (c) the change in the internal energy of the gas; (d) the thermal energy taken out of or put into the gas.

Short Answer

Expert verified
a) \( W = 2.40 \times 10^{6} \mathrm{J} \) b) \( T_f = 900.0 \mathrm{K} \) c) \( \Delta U \approx 60 \times 10^{3} \mathrm{J} \) d) \( Q \approx 2.46 \times 10^{6} \mathrm{J} \)

Step by step solution

01

- Understand the problem

We need to use the ideal gas law and thermodynamic relations to find the work done by the gas, the new temperature, the change in internal energy, and the thermal energy exchanged.
02

- Use the formula for work done by the gas

The work done by an ideal gas during expansion at constant pressure is given by \[ W = P \times \big( V_f - V_i \big) \], where \(P\) is the pressure, \(V_f\) is the final volume, and \(V_i\) is the initial volume.
03

- Calculate the work done by the gas

Substitute the given values into the formula: \(P = 6.00 \times 10^{6} \mathrm{Pa}\), \(V_i = 0.200 \mathrm{m}^3\), \(V_f = 0.600 \mathrm{m}^3\)\[ W = 6.00 \times 10^{6} \mathrm{Pa} \times (0.600 \mathrm{m}^3 - 0.200 \mathrm{m}^3) \]\[ W = 6.00 \times 10^{6} \mathrm{Pa} \times 0.400 \mathrm{m}^3 \]\[ W = 2.40 \times 10^{6} \mathrm{J} \]
04

- Find the temperature of the gas at the new volume using the ideal gas law

The ideal gas law is \( PV = nRT \). Since pressure is constant, \( V_i / T_i = V_f / T_f \). Rearrange to find \(T_f\):\[ T_f = T_i \times \frac{V_f}{V_i} \]Substitute the given values: \( T_i = 300.0 \mathrm{K}\), \( V_i = 0.200 \mathrm{m}^3 \), \( V_f = 0.600 \mathrm{m}^3 \)\[ T_f = 300.0 \mathrm{K} \times \frac{0.600 \mathrm{m}^3}{0.200 \mathrm{m}^3} \]\[ T_f = 300.0 \mathrm{K} \times 3 \]\[ T_f = 900.0 \mathrm{K} \]
05

- Determine the change in internal energy

For an ideal gas, the change in internal energy \( \Delta U \) is given by \( \Delta U = n C_V ( T_f - T_i ) \). Assuming a diatomic ideal gas \( C_V = \frac{5}{2} R \), and using \(n = \frac{PV}{RT} \):\[ n = \frac{6.00 \times 10^{6} \mathrm{Pa} \times 0.200 \mathrm{m}^3}{8.314 \mathrm{J \cdot K^{-1} \cdot mol^{-1}} \times 300.0 \mathrm{K}} \]\[ n \approx 4.82 \mathrm{mol}\]\[ \Delta U = 4.82 \mathrm{mol} \times \frac{5}{2} \times 8.314 \mathrm{J \cdot K^{-1} \cdot mol^{-1}} \times ( 900.0 \mathrm{K} - 300.0 \mathrm{K}) \]\[ \Delta U = 4.82 \times 20.785 \times 600 \]\[ \Delta U \approx 60 \times 10^{3} \mathrm{J} \]
06

- Determine the thermal energy exchanged

The thermal energy exchanged \(Q\) is given by \( Q = W + \Delta U \), where \(W\) is the work done by the gas and \(\Delta U\) is the change in internal energy:\[ Q = 2.40 \times 10^{6} \mathrm{J} + 60 \times 10^{3} \mathrm{J} \]\[ Q = 2.46 \times 10^{6} \mathrm{J} \]

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Thermodynamics
Thermodynamics is the branch of physics that deals with heat, work, and the forms of energy transformation. It plays an essential role in understanding how systems exchange energy and perform work. For instance, when we expand a gas at constant pressure, we use thermodynamic principles to compute work done by the gas and the changes in its internal energy. In the context of the ideal gas laws, thermodynamics helps us comprehend the behavior of gases under varying conditions of temperature, pressure, and volume. Recalling the first law of thermodynamics, which states that energy cannot be created or destroyed but only transferred or transformed, we can say that the change in internal energy of an ideal gas is the sum of the work done by the gas and the heat added to it.
Work Done by Gas
When an ideal gas expands or contracts, it performs work. The work done by a gas is a key concept in thermodynamics, and it is especially straightforward to calculate when the process occurs at a constant pressure. The formula for work done by the gas during expansion or compression at constant pressure is given by:

\ W = P \times ( V_f - V_i ) \

Here, \(P\) is the constant pressure, \(V_f\) is the final volume, and \(V_i\) is the initial volume. This relationship shows that work is the product of the pressure and the change in volume. In our example, we found that the work done by the gas during expansion from 0.200 \(\text{m}^3\) to 0.600 \(\text{m}^3\) at a pressure of 6.00 \( \times 10^{6} \) \(\text{Pa}\) is 2.40 \( \times 10^{6} \) \(\text{J}\).
Internal Energy
Internal energy is a measure of the total energy contained within a system, stemming from both kinetic and potential energies of its molecules. For an ideal gas, only the kinetic energy matters, as there are no intermolecular forces. The change in internal energy \(\Delta U\) of an ideal gas can be evaluated using the specific heat at constant volume \(C_V\):

\$ \Delta U = n C_V (T_f - T_i) \

\(n\) is the number of moles, \(C_V\) is the specific heat capacity at constant volume, and \((T_f - T_i)\) is the change in temperature. For a diatomic gas, \(C_V\) is \$ \frac{5}{2}R \ $$ where \(R\) is the universal gas constant (8.314 \(\text{J} \cdot \text{K}^{-1} \cdot \text{mol}\text{}^{-1}\)). In the given problem, we determined that the change in internal energy, \(\Delta U\), is roughly 6.0 \(\times 10^{4} \) \(\text{J}\).
Thermal Expansion
Thermal expansion of gases is the change in volume of a gas with a change in temperature at constant pressure. According to Charles’s Law, at constant pressure, the volume of an ideal gas is directly proportional to its absolute temperature. This can be mathematically represented as:

\ \ \$ \frac{V_i}{T_i} = \$ \frac{V_f}{T_f} \

This relationship allows us to determine the final temperature of the gas after expansion. When the initial and final volumes (\(V_i\) and \(V_f\)) and the initial temperature (\(T_i\)) are known, the final temperature (\(T_f\)) can be found by rearranging the equation:

\ T_f = T_i \$ \frac{V_f}{V_i} \$

In our exercise, the gas expanded from 0.200 \(\text{m}^3\) to 0.600 \(\text{m}^3\). Hence, the final temperature was found to be 900.0 K, tripling from the initial 300.0 K.

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Most popular questions from this chapter

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