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-1mol-1 , caculate Δ U and Δ pV for this process. (R = 8.0 JK-1mol-1]

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-1mol-1 , caculate Δ U and Δ pV for this process. (R = 8.0 JK-1mol-1] (A)  Δ U = 14 kJ: Δ(pV) = 4 kJ (B)  Δ U = 14 kJ: Δ(pV) = 18 kJ (C)  Δ U = 2.8 kJ: Δ(pV) = 0.8 kJ (D)  Δ U = 14 kJ: Δ(pV) = 0.8 kJ

Tardigrade Admin · Accepted Answer

Δ U=nCVm×Δ T = 5 × 28 × 100 = 14 kJ Δ PV=nRΔ T =5× 8× 100=4 kJ

Q. $5$ moles of an ideal gas at $100 K$ are allowed to undergo reversible compression till its temperature becomes $200 K$ . If $C_{V} = 28 J K^{- 1} m o l^{- 1}, c a c u l a t e Δ U$ and $Δ p V$ for this process. ( $R = 8.0 J K^{- 1} m o l^{- 1}$ ]

$Δ U = 14 k J : Δ (p V) = 4 k J$

$Δ U = 14 k J : Δ (p V) = 18 k J$

$Δ U = 2.8 k J : Δ (p V) = 0.8 k J$

$Δ U = 14 k J : Δ (p V) = 0.8 k J$

$Δ U = n C_{V} m \times Δ T$
$= 5 \times 28 \times 100 = 14 k J$
$Δ P V = n R Δ T$
$= 5 \times 8 \times 100 = 4 k J$

Q. 5 moles of an ideal gas at 100K are allowed to undergo reversible compression till its temperature becomes 200K. If CV​=28JK−1mol−1,caculateΔU and ΔpV for this process. (R=8.0JK−1mol−1]

ΔU=14kJ:Δ(pV)=4kJ

ΔU=14kJ:Δ(pV)=18kJ

ΔU=2.8kJ:Δ(pV)=0.8kJ

ΔU=14kJ:Δ(pV)=0.8kJ