Question

020 (part 1 of 2) 10.0 points An ideal gas initially is allowed to expand isothermally until its volume is 6.66667 L and ts pressure is 6 kPa. Then it is heated at constant volume until its pressure is 5 kPa Calculate the work done by the gas. Answer in units ofkJ 021 (part 2 of 2) 10.0 points Find the heat added to the gas during this process. Answer in units of kJ.

I can not find the answer to these questions. Everyone is multiplying liters with pascals and that is not how you get joules. I tried doing meters cubed times pascals but got the wrong answer. I thought it was an isochoric process as well so I thought work was zero but that was also wrong. If anyone knows how to do accurate thermodynamics, please help me!

It is NOT W=0 for the first part. I entered that and it was WRONG

initial volume is NOT missing this is all the information provided and I am expected to answer these questions with only the given information

Answer 1

020] V = V' = 6.66667 L = 0.00666667 m³

Work done in an isothermal process is given by:

020] V = V' = 6.66667 L = 0.00666667 m3 Work done in an isothermal process is given by W = n R T Ln(P_1/P_2) here, T can be found using PV = nRT taking n = 1, T = PV/R = 6000(0.006666667)/8.314 = 4.811 K so, work done by the gas is W = (8.314) (4.811) Ln (5000/6000) = -7.293 j From 1st law of thermodynamics, The heat added to the gas during this process = -Work done by the gas [since the change in internal energy = 0 J]therefore, heat added = + 7.293 J.

here, T can be found using PV = nRT

taking n = 1,

$\small T = \frac{PV}{R} = \frac{6000(0.00666667)}{8.314} = 4.811 K$

so, work done by the gas is:

$\small W = (8.314)(4.811)ln(5000/6000) = - 7.293 J$

020]

From 1st law of thermodynamics, The heat added to the gas during this process = - Work done by the gas

[since the change in internal energy = 0 J]

therefore, heat added = +7.293 J.