Question

5 moles of an ideal gas at 100 K are allowed to undergo reversible compression till its temperature becomes 200 K. If Cᵥ = 28 JK⁻¹ mol⁻¹, calculate ∆U and ∆pV for this process. (R = 8.0 J K⁻¹ mol⁻¹)
(A) ∆U = 2.8 kJ; ∆(pV) = 0.8 kJ (B) ∆U = 14kJ; ∆(pV) = 4kJ
(C) ∆U = 14kJ; ∆(pV) = 18kJ (D) ∆U = 14kJ; ∆(pV) = 0.8kJ

Answer 1

Answer:

Explanation:

Solution:- (A) ΔU=14kJ;Δ(pV)=4kJ

n=5;T

i

​

=100K;T

f

​

=200K;

C

v

​

=28J/molK; Ideal gas

ΔU=nC

v

​

ΔT

=5mol×28J/molK×(200−100)K

=14,000J=14kJ

⇒C

p

​

=C

v

​

+R=(28+8)J/molK

=36J/molK

⇒ΔH=nC

p

​

ΔT=5mol×36J/molK×100K

=18000J=18kJ

ΔH=ΔU+Δ(PV)

⇒Δ(PV)=ΔH−ΔU=(18−14)kJ=4kJ

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Answer 2

Answer:

5 moles of an ideal gas at 100 K are allowed to undergo reversible compression till its temperature becomes 200 K. If Cᵥ = 28 JK⁻¹ mol⁻¹, calculate ∆U and ∆pV for this process. (R = 8.0 J K⁻¹ mol⁻¹)

(A) ∆U = 2.8 kJ; ∆(pV) = 0.8 kJ (B) ∆U = 14kJ; ∆(pV) = 4kJ

(C) ∆U = 14kJ; ∆(pV) = 18kJ ™✓✓[

D) ∆U = 14kJ; ∆(pV) = 0.8kJ