Calculating ΔE Difference for 2 Samples of Monatomic Ideal Gas

[hellowmad]

Homework Statement

for Ideal gas under different pathway

Relevant Equations

E = Q+W
PV = nRT

Question: Two samples of a monatomic ideal gas are in separate containers at the same conditions of pressure, volume, and temperature (V = 1.00 L and P = 1.00 atm). Both samples undergo changes in conditions and finish with V = 2.00 L and P = 2.00 atm. However, in the first sample, the volume is changed to 2.0 L while the pressure is kept constant, and then the pressure is increased to 2.00 atm while the volume remains constant. In the second sample, the opposite is done. The pressure is increased first, with constant volume, and then the volume is increased under constant pressure. Calculate the difference in delta E between the first sample and the second sample.


for sample 1 is calculated by W = W1 +W2 = p1 delta(V) +0 = L atm, and for sample 2 W = 0+P2 deltaV = -2 L atm. But I have no idea how to calculate the Q for each samples.

 

[kuruman]

Are you aware that the change in internal energy is path-independent? The statement of the problem does not say that the samples have the same number of moles, but you may assume that they do.

 

[hellowmad]

got it. I forgot it. in that case the different is zero as both share the same initial and final conditions. Thank you! great help!

 

[DrClaude]

The statement of the problem does not say that the samples have the same number of moles, but you may assume that they do.

You can infer that the number of moles is the same from

Two samples of a monatomic ideal gas are in separate containers at the same conditions of pressure, volume, and temperature (V = 1.00 L and P = 1.00 atm).

 

[kuruman]

You can infer that the number of moles is the same from

That's what Avogadro said. I should have listened more carefully.