Recent content by Joseph95

Homework Statement A well-fitting piston with 4 small holes in a sealed water-filled cylinder,shown in Fig is pushed to the right at a constant speed of 4 mm /s while the pressure in the right compartment remains constant at 50 kPa gage. Disregarding the frictional effects,determine the force...

Thank you very much for your information and elaboration Sir.

Okay it is my fault Sir ! Sorry for that.Besides this could we obtain ΔH=ΔU+PΔV+VΔP by differentiating equation H=U+PV ?

We couldn't calculate the internal energy of a substance directly and we define it as change of state so we differentiate equation U=Q-PV and obtain dU=dQ-PdV and then integrating it as a result we get ΔU=ΔQ-PΔV.

Homework Statement I was trying to obtain Tds=dh-Vdp by differentiating the internal energy equation U=Q-PV and doing some arrangements but at the end I couldn't achieve my goal.The part that I don't understand is when we differentiate H=U+PV we obtain dH=dU + PdV + VdP and then from there we...

Thank you very much sir !

Qloss=Qgain m1Water.Cwater.ΔT1=m2water.Cwater.ΔT2 2 mole × 18g/mole × 1 cal/deg(100°C-T)=1 mole × 18g/mole × 1 cal/deg (T-0°C) 2(100°C-T)=T , 200°C=3T T≅66.7°C Calculation for 2 mole liquid water at 100°C Ti=100°C ------> 100+273.3=373.3°K Tf=66.7°C-------> 66.7 + 273.3=340°K ΔS=∫CpdT/T...

Ohh.You are right .I forget it.I am going to calculate again.

We should multiply entropy by the number of moles of each liquid before finding the net entropy of the system.Thus, If we mix 1 mole of liquid water at 0°C with 2 mole of liquid water at 100°C so the result should be: ΔS=1x3.02 cal/deg= 3.02 cal/deg ΔS=2×(-2.59 cal/deg)= -5.18 cal/deg...

ΔS=-2.59 X 1/2 cal/deg mol =-1.295 ΔS= 3.02 X 1/2 cal/deg mol =1.51 ΔS°=1.51-1.295=0.215 cal/deg mol I think this is the correct answer now.Is there any mistake in the unit of ΔS° ?

What if we mix 1 mole of liquid water at 0°C with 2 mole of liquid water at 100°C.How will the result change? Because in this equation ΔS=∫dQ/T entropy doesn't depend on quantity of substance.So can we say entropy doesn't depend on mole of the substance for the whole concept?

Step 1: Heat Loss=Heat Gain m.Cwater.ΔT1=m.Cwater.ΔT2 (Moles are equal so the mass) (100°C-T)=(T-0°C) T=50°C Step 2: Calculations for 0.5 mole liquid water at 100°C Ti=100°C----> 100 +273.3=373.3°K Tf=50°C------> 50+273.3=323.3°K ΔS=∫dQ/T pressure is constant therefore ΔS=∫CpdT/T from 373.3°K...

I am new in here.Forgive me.

Homework Statement Calculate the ΔS° when 0.5 mole of liquid water at 0°C is mixed with 0.5 mole of liquid water at 100°C.Assume Cp=18cal/deg mole over the whole range of temperatures. Homework Equations ΔS=∫ Cp/T dT