An ideal gas at a given state expands to a fixed final volume first at constant pressure and then at constant temperature. In which case is the work done greater?Group of answer choicesNone of themConstant volumeConstant temperatureConstant pressure

The work done by a gas is given by the area under the pressure-volume curve in a P-V diagram.

1. In the case of constant pressure (isobaric process), the work done is given by W=PΔV, where P is the pressure and ΔV is the change in volume. Since the volume is changing and the pressure is constant, there is work done by the gas.

2. In the case of constant temperature (isothermal process), the work done is given by W=nRTln(Vf/Vi), where n is the number of moles, R is the gas constant, T is the temperature, Vf is the final volume, and Vi is the initial volume. Since both the volume and the temperature are changing, there is also work done by the gas.

3. In the case of constant volume (isochoric process), there is no work done because the volume is not changing (ΔV=0), so the work done is W=PΔV=0.

Comparing the three processes, the work done is greatest in the isothermal process (constant temperature) if the final volume is larger than the initial volume. This is because the natural logarithm function (ln) in the work equation for the isothermal process can yield larger values than the constant pressure process, especially for large volume changes. Therefore, the work done is greater in the case of constant temperature.

The work done by a gas is given by the area under the pressure-volume curve in a P-V diagram.

1. In the case of constant pressure (isobaric process), the work done is given by W=PΔV, where P is the pressure and ΔV is the change in volume. Since the volume is changing and the pressure is constant, there is work done by the gas.

2. In the case of constant temperature (isothermal process), the work done is given by W=nRTln(Vf/Vi), where n is the number of moles, R is the gas constant, T is the temperature, Vf is the final volume, and Vi is the initial volume. Since the volume is changing and the temperature is constant, there is also work done by the gas.

3. In the case of constant volume (isochoric process), the work done is zero because there is no change in volume, and therefore no area under the P-V curve.

Comparing the isobaric and isothermal processes, the work done is generally greater in the isothermal process. This is because the pressure decreases as the gas expands, allowing for more work to be done.

So, the work done is greater in the case of constant temperature.