Answered

Get reliable answers to your questions at Westonci.ca, where our knowledgeable community is always ready to help. Explore a wealth of knowledge from professionals across various disciplines on our comprehensive Q&A platform. Our platform provides a seamless experience for finding reliable answers from a network of experienced professionals.

Two identical cylinders with a movable piston contain 0.7 mol of helium gas at a temperature of 300 K. The temperature of the gas in the first cylinder is increased to 564 K at constant volume by doing work W1 and transferring energy Q1 by heat. The temperature of the gas in the second cylinder is increased to 564 K at constant pressure by doing work W2 while transferring energy Q2 by heat.
A. Find ΔEint, 1, Q1, and W1 for the process at constant volume.
B. Find ΔEint, 2, Q2, and W2 for the process at constant pressure.

Sagot :

Answer:

Explanation:

A ) At constant volume :

ΔEint = n Cv x ΔT , n is no of moles , Cv is specific heat at constant volume , ΔT is increase in temperature .

For helium Cv = 3/2 R = 1.5 x 8.3 J = 12.45 J

ΔEint = .7 x 12.45 x ( 564 - 300 )

= 2300.76 J .

W₁ = 0 because volume is constant so work done by gas is zero .

Q₁ = ΔEint = 2300.76 J

B )

At constant pressure

Q₂ = n Cp Δ T , Cp is specific heat at constant pressure .

For monoatomic gas ,

Cp = 5/2 R = 2.5 x 8.3 J = 20.75 J

Q₂ = .7 x 20.75 x 264 J

= 3834.6 J

W₂ = work done by gas

= PΔV = nRΔT

= .8  x 8.3 x 264

= 1752.96 J

ΔEint = Q₂ - W₂

= 3834.6 - 1752.96

= 2081.64 J.

okpalawalter8

 ΔEint, 1, Q1, and W1 for the process at constant volume. and ΔEint, 2, Q2, and W2 for the process at constant pressure is mathematically given as

a)

dE1= 2300.76 J .

W1=0 as Volume is constant

Q1= 2300.76 J as Q= dE1

b)

Q2= 3834.6 J

W2= 1752.96 J

dE2= 2081.64 J.  

What is ΔEint, Q1 and W1 for the process at constant volume and pressure?

a)

Generally, the equation for the Constant Volume   is mathematically given as

dE = n Cv x dT

Where

Cv = 3/2

R = 1.5 x 8.3 J

R= 12.45 J

Therefore

dE = 0.7 x 12.45 x ( 564 - 300 )

dE1= 2300.76 J .

W1=0 as Volume is constant

Q1= 2300.76 J as Q= dE1

b)

Generally

Q2 = n Cp dT

Where

Cp = 5/2

R = 2.5 x 8.3 J

R= 20.75 J

Hemce

Q2 = 0.7 x 20.75 x 264 J

Q2= 3834.6 J

For Work done

W=PdV

W= nRdT

Therefore

W= 0.8  x 8.3 x 264

W2= 1752.96 J

Hence

dE = Q₂ - W₂

dE= 3834.6 - 1752.96

dE2= 2081.64 J.  

For more information on Pressure

https://brainly.com/question/25688500

Thanks for using our platform. We're always here to provide accurate and up-to-date answers to all your queries. Thanks for stopping by. We strive to provide the best answers for all your questions. See you again soon. Get the answers you need at Westonci.ca. Stay informed by returning for our latest expert advice.