One mole of a monatomic ideal gas undergoes an adiabatic expansion in which its volume becomes eight times its initial value. If the initial temperature of the gas is and the universal gas constant the decrease in its internal energy, in Joule, is _____

Question

One mole of a monatomic ideal gas undergoes an adiabatic expansion in which its volume becomes eight times its initial value. If the initial temperature of the gas is $100 K$ and the universal gas constant $R = 8.0 {J mol}^{- 1} K^{- 1},$ the decrease in its internal energy, in Joule, is ________. [2018]

Smart Achievers · Accepted Answer

(900)

$Given: T_{i} = 100 K, V_{f} = 8 V_{i}$

For an adiabatic process, $T V^{γ - 1} = constant$

or, $T_{i} V_{i}^{γ - 1} = T_{f} V_{f}^{γ - 1}$

$\Rightarrow \frac{T_{i}}{T_{f}} = {(\frac{V_{f}}{V_{i}})}^{γ - 1}$ $\Rightarrow \frac{T_{i}}{T_{f}} = {(\frac{8 V_{i}}{V_{i}})}^{γ - 1}$

For monoatomic gas, $γ = \frac{5}{3}$

$∴$ $T_{f} = \frac{T_{i}}{8^{(\frac{5}{3} - 1)}} = \frac{T_{i}}{4}$

Change in internal energy, $Δ u = n C_{v} Δ T$

$= 1 \times \frac{3}{2} R (\frac{T_{i}}{4} - T_{i})$ $= \frac{3}{2} \times 8 \times (\frac{- 3}{4}) \times 100 = - 900 J$

Solution

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Solution

Q. One mole of a monatomic ideal gas undergoes an adiabatic expansion in which its volume becomes eight times its initial value. If the initial temperature of the gas is 100 K and the universal gas constant R=8.0 J mol-1K-1, the decrease in its internal energy, in Joule, is ________. [2018]

Ans. (900) Given: Ti=100 K, Vf=8Vi For an adiabatic process, TVγ-1=constant or, TiViγ-1=TfVfγ-1 ⇒TiTf=(VfVi)γ-1⇒TiTf=(8ViVi)γ-1 For monoatomic gas, γ=53 ∴ Tf=Ti8(53-1)=Ti4 Change in internal energy, Δu=nCvΔT =1×32R(Ti4-Ti)=32×8×(-34)×100=-900 J

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