# Archived Finding the thermal energy and Q for an isobaric process.

#### Brittbrit22

1. Homework Statement

How much work is done on the gas in the process shown in figure P12.24. This is from College Physics: A strategic approach, 2nd addition, by Knight, et al. Ch.12, #24

The graph shown is kPa vs V(cm^3). It's a constant temperature process (isobaric) at 200 kpa. I was able to answer the question posed in the textbook. My answer was .0004 J for the amount of work. I am not sure if the work should be negative or not. Also, I do not have access to the correct answer so I do not know if this value is correct either.

My professor also asked up to calculate the ΔEth or change in internal thermal energy and the Q value for this process. I do understand how these can be calculated if I was given no information about the initial or final temperatures or the number of moles. I can't calculate the number of moles with out a temperature and I can't calculate Q without the number of moles. The problem states "a gas" but if I knew what gas I could just use the molar mass to find moles but they don't specify what gas it is.

2. Homework Equations

1. Wgas=pΔV

2. ΔEth=Q - Wgas

3. n=pv/rt

4. Q=nCpΔt

3. The Attempt at a Solution

I tried to use equation 2 and I would just solve for Q. But, I can't figure out Q if I can't calculate the delta Eth first. I can't calculate the Eth because I do no know the number of moles. Then I tried to use equation 3 to find the number of moles (n) but I do not know what t(t for temperature in kelvin) is.

So n=0.2pa*0.0001m^3/ (8.31J/mol.k * t) but I don't know t, and I don't see how I can find it without moles.

Then Q=n*20.8*t (I do not know the values for n(moles) or t(temperature).

Or if I were to solve for Q in equation 2 : 3/2*8.31 J/mol.k*t=Q-(-0.0004J) >>> But again, I do not know t(the temperature in kelvin) nor I understand how I am supposed to find that value.

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#### Chestermiller

Mentor
The OP, in his problem statement, says that the temperature is constant. This means that, for a (presumed) ideal gas, the change in internal energy is zero. Therefore, the heat is equal to the work.