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1,250.0 moles of an ideal at a temperature of -11.0 degrees Celsius is contained in a volume of 60.0 L and then expands to 110.0 L at constant temperature. How much work was done by the gas in this expansion?

User Reigo Hein
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1 Answer

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Final answer:

The work done by the gas in the expansion is -62.5 atm*L.

Step-by-step explanation:

The work done by a gas in an expansion can be calculated using the formula:

Work = -PΔV

Where P is the pressure and ΔV is the change in volume. In this case, the gas expands from 60.0 L to 110.0 L at constant temperature.

The change in volume is

110.0 L - 60.0 L = 50.0 L.

The work done by the gas can be calculated as:

Work = -PΔV

= -(1.2500 atm)(50.0 L)

= -62.5 atm*L

User Grzegorz Smulko
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