Question

find the internal energy of two moles of a diatomic gas at 300k.​

Answer 1

Answer:

Given that:-

P

ext.

=1atm

P

1

=5atm

P

2

=2atm

T

1

=300K

T

2

=T(say)=?

n=2

C

V

=

2

5

R

For an adiabatic process,

W=nC

V

dT=−PΔV

∴nC

V

(T

2

−T

1

)=−P

ext

(V

2

−V

1

)

⇒2×

2

5

R(T−300)=−1(

P

2

nRT

2

−

P

1

nRT

1

)

⇒5R(T−300)=−2R(

2

T

−

5

300

)

⇒5T−1500=−T+120

⇒6T=1620⇒T=270K

∴T

2

=T=270K

∴W=nC

V

dT=2×

2

5

R×(270−300)=−1247.1J

⇒q=0

ΔU=W=−1247.1J

∴ΔH=ΔU+nRΔT

⇒ΔH=−1247.1+2×8.314×(270−300)=−1745.94J

Explanation:

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Answer 2

The internal energy of two moles of diatomic gas at 300k is 1745.94 J

Given:

Diatomic gas at 300k

Solution:

Internal Energy: Internal energy encompasses all of the energy contained within a given system, including kinetic energy of molecules and energy stored in all chemical bonds between molecules. Every time a system is changed, there are energy transfers and conversions due to the interactions of heat, work, and internal energy.

We need to the internal energy of two moles of a diatomic gas at 300k,

For a diatomic gas,

P1= 5 atm, P2 = 2 atm

T1 = 300k

Cv = 5/2 R

P(ext) = ⊥

n =2

For a Adiabatic process,

W = n*Cv(T2 - T1) = - P(ext)$(\frac{nRT2}{P2} - \frac{nRT1}{P1} )$

⇒ 2*5/2 R (T2 - 300) = -⊥*nR$(\frac{T2}{P2}- \frac{T1}{P1} )$

⇒ 5(T2 - 300) = -2*$(\frac{5T2 - 600}{10})$

⇒ 5T2 - 1500 = -2*5( T2 - 120/10)

⇒ 5T2 - 1500 = -10(T2 - 120/10)

⇒ 5T2 - 1500 = -T2 + 120

≅ 6T2 = 1620 ≡ T2 = 270K

From, W = n*Cv(T2 - T1)

⇒2*5/2 R*(-30)

⇒5*8.314*(-30)

≅ -1247.1 J

q = 0

ΔV = W

ΔH= ΔV+nRΔT

= -1247.1 + 2*8.314*(-30)

⇒ 1745.94 J

∴ The internal energy of two moles of diatomic gas at 300k is 1745.94 J.

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