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### Isothermal

Posted: Thu Feb 06, 2020 8:22 pm
What exactly does it mean for the expansion of a gas to be “isothermal”? How does this impact which equations we should use?

### Re: Isothermal

Posted: Thu Feb 06, 2020 8:28 pm
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.

### Re: Isothermal

Posted: Thu Feb 06, 2020 8:38 pm
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.

### Re: Isothermal

Posted: Thu Feb 06, 2020 10:00 pm
isothermal refers to the system being at constant temperature.

### Re: Isothermal

Posted: Thu Feb 06, 2020 10:46 pm
Isothermal refers to the constant temperature of a system. No heat is added or lost to the environment.

### Re: Isothermal

Posted: Fri Feb 07, 2020 10:50 am
Isothermal means that delta T = 0 and therefore delta U = nc(delta T) = 0. This does not mean that delta Q = 0.

### Re: Isothermal

Posted: Fri Feb 07, 2020 12:29 pm
Isothermal means that there is a constant temperature and that delta U is also equal to 0

### Re: Isothermal

Posted: Fri Feb 07, 2020 1:22 pm
If the expansion of the gas is isothermal, then this means the temperature remains constant and delta U = 0

### Re: Isothermal

Posted: Fri Feb 07, 2020 1:28 pm
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.

### Re: Isothermal

Posted: Fri Feb 07, 2020 2:53 pm
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.

### Re: Isothermal

Posted: Fri Feb 07, 2020 3:08 pm
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

### Re: Isothermal

Posted: Fri Feb 07, 2020 3:46 pm
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

### Re: Isothermal

Posted: Fri Feb 07, 2020 3:59 pm
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

### Re: Isothermal

Posted: Sat Feb 08, 2020 12:34 pm
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?

### Re: Isothermal

Posted: Sat Feb 08, 2020 1:37 pm
An isothermal process is a change of system where the temperature remains constant. Therefore, delta U would be equal to 0.

### Re: Isothermal

Posted: Sun Feb 09, 2020 10:34 pm
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

### Re: Isothermal

Posted: Sun Feb 09, 2020 10:44 pm
along with isothermal, i think we also need to isobaric (constant pressure) and isochoric (constant volume)

### Re: Isothermal

Posted: Mon Feb 10, 2020 11:48 am
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)