## Isothermal

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

Fdonovan 3D
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### Isothermal

What exactly does it mean for the expansion of a gas to be “isothermal”? How does this impact which equations we should use?

Rachel Yu 1G
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### Re: Isothermal

The expansion of a gas is isothermal when the volume of a system changes with a constant temperature. It is best to use equations where volume and pressure change rather than temperature change.

Michael Nguyen 1E
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### Re: Isothermal

The word "isothermal" refers to a system being at a constant temperature. In these situations, it is best to use the equations that involve a change in volume and pressure.

205291012
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### Re: Isothermal

isothermal refers to the system being at constant temperature.

Jessica Katzman 4F
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### Re: Isothermal

Isothermal refers to the constant temperature of a system. No heat is added or lost to the environment.

Sean Cheah 1E
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### Re: Isothermal

Isothermal means that delta T = 0 and therefore delta U = nc(delta T) = 0. This does not mean that delta Q = 0.

Kallista McCarty 1C
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### Re: Isothermal

Isothermal means that there is a constant temperature and that delta U is also equal to 0

Celine 1F
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### Re: Isothermal

If the expansion of the gas is isothermal, then this means the temperature remains constant and delta U = 0

Abby Soriano 1J
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### Re: Isothermal

Celine 1F wrote:If the expansion of the gas is isothermal, then this means the temperature remains constant and delta U = 0

Could you explain why delta U would equal 0? I'm a little confused about the connection between temperature and delta U.

CameronDis2K
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### Re: Isothermal

Isothermal means constant temperature, or the change in temperature = 0. For the work equations, the irreversable expansion equation can be used, which is W = - P external * deltaV.

MeeraBhagat
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### Re: Isothermal

Abby Soriano 1J wrote:
Celine 1F wrote:If the expansion of the gas is isothermal, then this means the temperature remains constant and delta U = 0

Could you explain why delta U would equal 0? I'm a little confused about the connection between temperature and delta U.

Delta U = 0 because the temperature of the system is not changing, meaning that the energy lost from the system through doing work is regained by energy from heat which is where we get the equation q = -w

Lizette Noriega 1H
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### Re: Isothermal

Isothermal means there is a constant temperature present in the system, therefore, the most ideal equations to use would be the ones that include a change in volume or pressure

Emily Chirila 2E
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### Re: Isothermal

Abby Soriano 1J wrote:
Celine 1F wrote:If the expansion of the gas is isothermal, then this means the temperature remains constant and delta U = 0

Could you explain why delta U would equal 0? I'm a little confused about the connection between temperature and delta U.

We know that delta U= q + w and delta U=(3/2)nR(deltaT)
if delta T=0 then you are multiplying the constants in the 2nd equation by 0, thus you end up with delta U= 0

if delta U= 0 then you can plug the zero into U= q + w to get:
0= q + w
therefore q=-w

Posts: 125
Joined: Sat Aug 17, 2019 12:17 am

### Re: Isothermal

Emily Chirila 2E wrote:
Abby Soriano 1J wrote:
Celine 1F wrote:If the expansion of the gas is isothermal, then this means the temperature remains constant and delta U = 0

Could you explain why delta U would equal 0? I'm a little confused about the connection between temperature and delta U.

We know that delta U= q + w and delta U=(3/2)nR(deltaT)
if delta T=0 then you are multiplying the constants in the 2nd equation by 0, thus you end up with delta U= 0

if delta U= 0 then you can plug the zero into U= q + w to get:
0= q + w
therefore q=-w

I understand the process above, however, isn't q = cmdeltaT, so as it is isothermic, delta T is equal to zero, so q is equal to zero?

Jessica Kwek 4F
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### Re: Isothermal

An isothermal process is a change of system where the temperature remains constant. Therefore, delta U would be equal to 0.

Nick Lewis 4F
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### Re: Isothermal

Jessica Kwek 4F wrote:An isothermal process is a change of system where the temperature remains constant. Therefore, delta U would be equal to 0.

I agree with this, for example, in pizza rolls review session they showed how when t = o, delta U = 0, therefore q = -w, which can come in useful knowing some of the types of questions that would be on the midterm

Zoya Mulji 1K
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### Re: Isothermal

along with isothermal, i think we also need to isobaric (constant pressure) and isochoric (constant volume)

Emily Chirila 2E
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### Re: Isothermal

Emily Chirila 2E wrote:
Abby Soriano 1J wrote:
Could you explain why delta U would equal 0? I'm a little confused about the connection between temperature and delta U.

We know that delta U= q + w and delta U=(3/2)nR(deltaT)
if delta T=0 then you are multiplying the constants in the 2nd equation by 0, thus you end up with delta U= 0

if delta U= 0 then you can plug the zero into U= q + w to get:
0= q + w
therefore q=-w

I understand the process above, however, isn't q = cmdeltaT, so as it is isothermic, delta T is equal to zero, so q is equal to zero?

If there is no change in temp, you calculate q using q=m(deltaH)

Think about the heating curve graph-- where there is a horizontal line (phase change) we use q=m(deltaH), whereas when there is a change in temp, we use the equation you stated: q=cm(deltaT)