Answer to Question #83420 in Molecular Physics | Thermodynamics for Naima

Question

6. A system absorbs 800J heat energy from the environment and its internal energy increase
by 500J. Find the work done by the system on the environment.
7. A Carnot’s engine has the same efficiency between 1000 k and 500 k between X k and
1000 k (this being the temperature of the sink in this case). Calculate the value of X.
8. A quantity of gas a cylindrical while doing work of 360 J on the environment absorbs
heat of 955J. What is the change in internal energy of the gas?
9. 1 mole gas is initially at 0ᵒ C. It is allowed to expand at constant pressure to double its
volume. How much heat will be required? Given C v =20.7Jmol -1 K -1 and R=8.31Jmol -1 K -1 .
10. At what temperature both the Farenheit and Kelvin scales give the same value?

Answer 1

7. A Carnot’s engine has the same efficiency between 1000 K and 500 K between X K and 1000 K (this being the temperature of the sink in this case). Calculate the value of X.

Solution:

η_1=1-500/1000=5/10

η_2=1-1000/x

Since

η_1=η_2

5/10=1-1000/x

1000/x=1-5/10=5/10

x=(1000×10)/5=2000 K

Answer: 2000 K

10. At what temperature both the Farenheit and Kelvin scales give the same value?

Solution:

tF=1.8tC+32.

tC=T−273.15.

tF=1.8T−459.67.

Since we want the Fahrenheit and Kelvin values equal

tF=T

T=1.8T−459.67

0.8T=459.67 00

T=459.67/0.8=574.5875

574.5875 F = 574.5875 K

Answer: 574.5875

8. A quantity of gas a cylindrical while doing work of 360 J on the environment absorbs heat of 955J. What is the change in internal energy of the gas?

Solution:

Q=∆E+W

∆E=Q-W

∆E=955 J-360 J=595 J

Answer: 595 J

6. A system absorbs 800J heat energy from the environment and its internal energy increase by 500J. Find the work done by the system on the environment.

Solution:

Q=∆E+W

W=Q-∆E

W=800 J-500 J=300 J

Answer: 300 J

9. 1 mole gas is initially at 0ᵒ C. It is allowed to expand at constant pressure to double its volume. How much heat will be required? Given C v =20.7Jmol -1 K -1 and R=8.31Jmol -1 K -1 .

Solution:

First Law of Thermodynamics: ΔU = n CV ΔT = Q - W

Q = n CV ΔT + W

W = p ΔV = p(V2 - V1)

p V2 = n R T2

p V1 = n R T1

W = p(V2 - V1) = n RΔT

Q = n CV ΔT + n RΔT = n (CV + R ) ΔT

Q = 1 x (20.7 + 8.31) x 273 = 4.7 x 103 J

Answer: 4.7 x 103 J

7. A Carnot’s engine has the same efficiency between 1000 K and 500 K between X K and 1000 K (this being the temperature of the sink in this case). Calculate the value of X.

Solution:

η_1=1-500/1000=5/10

η_2=1-1000/x

Since

η_1=η_2

5/10=1-1000/x

1000/x=1-5/10=5/10

x=(1000×10)/5=2000 K

Answer: 2000 K

10. At what temperature both the Farenheit and Kelvin scales give the same value?

Solution:

tF=1.8tC+32.

tC=T−273.15.

tF=1.8T−459.67.

Since we want the Fahrenheit and Kelvin values equal

tF=T

T=1.8T−459.67

0.8T=459.67 00

T=459.67/0.8=574.5875

574.5875 F = 574.5875 K

Answer: 574.5875

8. A quantity of gas a cylindrical while doing work of 360 J on the environment absorbs heat of 955J. What is the change in internal energy of the gas?

Solution:

Q=∆E+W

∆E=Q-W

∆E=955 J-360 J=595 J

Answer: 595 J

6. A system absorbs 800J heat energy from the environment and its internal energy increase by 500J. Find the work done by the system on the environment.

Solution:

Q=∆E+W

W=Q-∆E

W=800 J-500 J=300 J

Answer: 300 J

9. 1 mole gas is initially at 0ᵒ C. It is allowed to expand at constant pressure to double its volume. How much heat will be required? Given C v =20.7Jmol -1 K -1 and R=8.31Jmol -1 K -1 .

Solution:

First Law of Thermodynamics: ΔU = n CV ΔT = Q - W

Q = n CV ΔT + W

W = p ΔV = p(V2 - V1)

p V2 = n R T2

p V1 = n R T1

W = p(V2 - V1) = n RΔT

Q = n CV ΔT + n RΔT = n (CV + R ) ΔT

Q = 1 x (20.7 + 8.31) x 273 = 4.7 x 103 J

Answer: 4.7 x 103 J