# Can anyone help me with this thermochemistry problem?

• michealyap
In summary, the student attempted to find the work done to lift a platform of weight 900 kg up a height of 2.5 meters using an isothermal process equation. They claimed they solved the first question but got an incorrect answer for the second. They also wanted to solve for the mass in the room and in the tank at the initial and final states, but were not successful.

area=2.25
p1=1bar
v1=0.23m3
p2=
v2=1m3

## Homework Equations

Isothermal process W=nrt ln(v2/v1)
W=F*D

## The Attempt at a Solution

I got the workdone then sub into the equation of isothermal process. i couldn't get the answer right

#### Attachments

• WhatsApp Image 2018-10-14 at 21.38.27 (2).jpeg
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Good morning from CEST. Thanks for posting your question to the right forum and using the template. However, we cannot see what you've achieved so far and what you've done, so please show your work or explain your thougts more accurately. The picture you attached shows two questions to solve, a) and b). Have you solved one of them or are you stuck on the first one? By the way, you don't need to calculate the work done to find the answers.

I have already solved the first question. a) i take the formula of P=F/A the total weight of is 900kg, to get the F, i multiplied 900kg with the gravity acceleration,9.81 m/s2 , area is given, 2.25m2 . Then substitute into it, i got 1.039bar.

For question b, i need to calculate the p2 to maintain the lift in the mentioned height (2.5m) . First i get the work done first W= F*D , after i get the workdone, i substitute the value into the isothermal process equation W=nRT in(v2/v1) , by comparison with the another equation, W=nRT In (p1/p2) then i can get for p2. But the answer i got was incorrect.

You neither know the ##R## nor ##T## nor ##n## (or ##m##), so you introduce more variables than you obtain equations. If I've understood correctly, you want to solve

$$mg\Delta z = nRTln\left(\frac{p_1}{p_2}\right) = nRTln\left(\frac{v_2}{v_1}\right)$$.

What result did you obtain?

Try to solve it with state variables only. Do you know the correlation between pressure ##p## and volume ##V## for an isothermal process ##T=const.##?

I understand the relationship of pressure p and volume V for isothermal process. P1V1 = P2V2. In the question i know V1,V2, and P1 i take as 1.039bar. Then ended up P2 is a wrong answer.

michealyap said:
I understand the relationship of pressure p and volume V for isothermal process. P1V1 = P2V2. In the question i know V1,V2, and P1 i take as 1.039bar. Then ended up P2 is a wrong answer.

That's because you don't take into account mass conservation, but in ##V_2## will be more mass than in ##V_1##, since molecules will be transferred from the tank. Can you express the mass (or moles) contained in the tank and in the room below the platform for the initial and the final state by means of ##T##, ##p##, ##R## and ##V##?

I am confused now. Initial state, does it mean that the gas in the tank and in the room is already mixed but the lift is just about to be lifted?

if that is the case mass in the room + mass in the tank = p(eqilibrium) *V / RT . but what is the R in here?

Mass in room + mass in tank = Pr1*Vr1 / RT + Pt1*Vt1 /RT - Initial state.
Mass in room + mass in tank = Pr2*Vr2 /RT + Pt2*Vt2 /RT - final state

At here we know Pr1,Vr1, Vt1, but we don't know Pr2 and Pt2 and Pt1.

michealyap said:
Mass in room + mass in tank = Pr1*Vr1 / RT + Pt1*Vt1 /RT - Initial state.
Mass in room + mass in tank = Pr2*Vr2 /RT + Pt2*Vt2 /RT - final state

$$m_{t\,1}=\frac{p_{t\,1}V_{t\,1}}{RT}$$
$$m_{r\,1}=\frac{p_{r\,1}V_{r\,1}}{RT}$$
$$m_{t\,2}=\frac{p_{t\,2}V_{t\,2}}{RT}$$
$$m_{r\,2}=\frac{p_{r\,2}V_{r\,2}}{RT}$$
mass conservation gives us $$m_{t\,1}+m_{r\,1}=m_{t\,2}+m_{r\,2}$$

michealyap said:
At here we know Pr1,Vr1, Pt1, Vt1, but we don't know Pr2 and Pt2 and also Vr2 and Vt2, but we know the Vr2+Vt2 =

I disagree: we know ##p_{r\,2}## and ##p_{t\,2}##.

Yes, i also agree we don't know Pr2 and Pt2 which is what we want to find. we don't know pt1 as well. Even with those equations we can not find three unknow with only two equations.

michealyap said:
Yes, i also agree we don't know Pr2 and Pt2 which is what we want to find. we don't know pt1 as well. Even with those equations we can not find three unknow with only two equations.

We DO know ##p_{r\,2}## and ##p_{t\,2}##. First of all, they must be equal, after the platform was elevated by ##\Delta z##. Secondly, the weight of the persons on the platform didn't change, neither did its area. The force applied by them on the platform must be identical.

yeah. we still unable to solve it right, we just know it is equal. the equation will become

Pr1*Vr1 + Pt1*Vt1 =Pr2*Vr2 + Pt2*Vt2 , let Pt2 = Pr2 = Peq,

Pr1*Vr1 + Pt1*Vt1 =Peq*Vr2 + Peq*Vt2 , we do not know the Pt1 and Peq.

michealyap said:
yeah. we still unable to solve it right, we just know it is equal. the equation will become

Pr1*Vr1 + Pt1*Vt1 =Pr2*Vr2 + Pt2*Vt2 , let Pt2 = Pr2 = Peq,

Pr1*Vr1 + Pt1*Vt1 =Peq*Vr2 + Peq*Vt2 , we do not know the Pt1 and Peq.

Then let's solve that question first: What pressure is necessary to elevate people who weigh 600 kg? That should give us ##p_{eq}##.

a) How i do it P=FA = 900kg * 9.81/2.25 = 3924 Pa since Pr1 is 1.0 bar the and the elevator is not lifted. so the pressure necessary to elevate is 1.039 bar. which i presume is Pr1.

michealyap said:
a) How i do it P=FA = 900kg * 9.81/2.25 = 3924 Pa since Pr1 is 1.0 bar the and the elevator is not lifted. so the pressure necessary to elevate is 1.039 bar. which i presume is Pr1.

So, if the weight of people and platform and the area of the platform don't change, why should a different pressure be needed to compensate their weight? ##p_{r\,1}=p_{eq}##

Yes.. You are right. Wao.. This is amazing.. I thought when it reached equilibrium the tank will also have the same pressure as the room..but why we can assume pt2 to be same in the room pr2.

michealyap said:
Yes.. You are right. Wao.. This is amazing.. I thought when it reached equilibrium the tank will also have the same pressure as the room..but why we can assume pt2 to be same in the room pr2.

The tank and the room will have the same pressure, as you stated two posts (#12) earlier: ##p_{t\,2}=p_{r\,2}=p_{eq}##. Maybe we have a different interpretation of the questions. I read it like this:

1) There is a platform with people on it, below the platform there is a reservoir of ideal gas. There is a equilibrium af forces between the weight (of people and platform) and the pressure in the reservoir. Calculate the pressure in the reservoir.

2) The reservoir is connected to a tank via a pipe with a valve, initially the valve is closed. After opening the valve the platform is elevated by the incoming ideal gas, before the equilibrium is established again. Assuming an isothermal state of change of the ideal gas, calculate the necessary (initial) pressure in the tank to lift the platform by 2.5 m.

michealyap said:
v1=0.23m3
p2=
v2=1m3
Please take more care in writing your equations. At the very least, indicate exponents by using ^, as in v1 = .23m^3 and v2 = 1 m^3.

Better yet, use the ##x_2## and ##x^2## icons on the menu bar to write subscripts and exponents, like this:
v1 = 0.23m3

For 2) I interpreted as what is the pressure need to be maintained , but not to elevate the lift.

Based on your interpretation, I fully understand what we have been.doing..

michealyap said:
Based on your interpretation, I fully understand what we have been.doing..

... and I'm confident that my interpretation is correct, since the result matches the one of your professor.