## Isothermal expansion of an ideal gas

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

Hannah Lee 2F
Posts: 117
Joined: Thu Jul 11, 2019 12:15 am

### Isothermal expansion of an ideal gas

For the isothermal expansion of an ideal gas in which deltaU = 0, is heat energy still being transferred into or out of the system and why?

Selena Yu 1H
Posts: 108
Joined: Fri Aug 09, 2019 12:16 am

### Re: Isothermal expansion of an ideal gas

Heat energy is still being transferred into or out of the system if work is being done on or by the system because in order for deltaU to be equal to zero then there must be heat energy to balance out the amount of work done for the deltaU = 0 because deltaU = heat energy + work. However, if no work is done and it is constant volume then heat energy is not being transferred into or out of the system.

005324438
Posts: 51
Joined: Sat Aug 24, 2019 12:16 am

### Re: Isothermal expansion of an ideal gas

Assuming it's still the expansion of an ideal gas, heat will have to be entering the system. According to Charles law, we know the volume of a gas is inversely proportional to its temperature, so for a gas to increase in volume but stay the same temperature, heat energy must be entering it.