6th edition, 8.9

$w=-P\Delta V$
and
$w=-\int_{V_{1}}^{V_{2}}PdV=-nRTln\frac{V_{2}}{V_{1}}$

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Tyra Nguyen 4H
Posts: 74
Joined: Fri Sep 28, 2018 12:25 am

6th edition, 8.9

The question is as follows:

8.9 An ideal gas in a cylinder was placed in a heater and gained 5.50 kJ of energy as heat. If the cylinder increased in volume from 345 mL to 1846 mL against an atmospheric pressure of 750. Torr during this process, what is the change in internal energy of the gas in the cylinder?"

I converted the mL to meters to use the work equation, but am not sure of what else to do.

Bianca Barcelo 4I
Posts: 74
Joined: Fri Sep 28, 2018 12:17 am
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Re: 6th edition, 8.9

The internal energy (U) = heat + work (q+w).

You would have to use the work equation (w=-P(deltaV)). So you would have to change the pressure from torr to atm and then find the change in volume from the given information. After getting the value of work, you would have to change that to Joules so use the conversion factor (101.325J/L*atm).

We are already given the value q, so just add q (5.50kj) and w together to find the total change in internal energy.

I hope that helps!

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