A 20.0-L volume of an ideal gas in a cylinder with a piston is at a pressure of 3.0 atm. Enough weight is suddenly removed from the piston to lower the external pressure to 1.5 atm. The gas then expands at constant temperature until its pressure is 1.5 atm. Find w.

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Identify the type of process: The problem describes an isothermal expansion of an ideal gas, where the temperature remains constant.

Use the formula for work done by an ideal gas during isothermal expansion: \( w = -nRT \ln\left(\frac{V_f}{V_i}\right) \). However, since we don't have \( n \) or \( T \), we can use the pressure-volume relationship for isothermal processes.

Apply the ideal gas law to relate initial and final states: \( P_iV_i = P_fV_f \) since \( nRT \) is constant.

Rearrange the equation to find the final volume \( V_f \): \( V_f = \frac{P_iV_i}{P_f} \).

Calculate the work done using the formula for work in terms of pressure and volume change: \( w = -P_{ext} \Delta V \), where \( \Delta V = V_f - V_i \) and \( P_{ext} = 1.5 \text{ atm} \).

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주요 개념

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Ideal Gas Law

The Ideal Gas Law relates the pressure, volume, temperature, and number of moles of an ideal gas through the equation PV = nRT. This law is fundamental in understanding gas behavior under varying conditions, such as changes in pressure and volume, which are central to the question regarding the gas expansion.

Work Done by a Gas

In thermodynamics, the work done by a gas during expansion or compression can be calculated using the formula w = -P_extΔV, where P_ext is the external pressure and ΔV is the change in volume. This concept is crucial for determining the work done when the gas expands against a lower external pressure.

Isothermal Process

An isothermal process occurs at a constant temperature, meaning that the internal energy of an ideal gas remains unchanged during expansion or compression. In this scenario, since the gas expands isothermally, the temperature remains constant, which influences how we calculate the work done and the relationship between pressure and volume.