One mole of an ideal monoatomic gas is heated at a constant pressure from 0∘C to 100∘C. Then the change in the internal energy of the gas is (Given, R=8.32Jmol−1K−1 )

Given that T1=0°C=273K,T2=100°C=373KFor monoatomic n = 1We know that Δ=ΔU=nCvT=3/2nRT =3/2×8.32×100 =3×8.32×50JΔU=1248J =1.25×103J

One mole of an ideal monoatomic gas is heated at a constant pressure from 0∘C to 100∘C. Then the change in the internal energy of the gas is (Given, R=8.32Jmol−1K−1 )

Given that T1=0°C=273K,T2=100°C=373KFor monoatomic n = 1We know that Δ=ΔU=nCvT=3/2nRT =3/2×8.32×100 =3×8.32×50JΔU=1248J =1.25×103J

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One mole of an ideal monoatomic gas is heated at a constant pressure from $0^{\circ} C$ to $100^{\circ} C .$ Then the change in the internal energy of the gas is (Given, $R = 8.32 J m o l^{- 1} K^{- 1}$ )

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$2.08 \times 10^{3} J$

$1.25 \times 10^{3} J$

$0.83 \times 10^{3} J$

$4.6 \times 10^{3} J$

answer is D.

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Detailed Solution

Given that $T_{1} = 0 ° C = {273K,T}_{2} = 100 ° C = 373K$

For monoatomic n = 1

We know that

$\begin{array}{l} Δ = Δ U = n CvT = 3/2nRT = 3/2 \times 8.32 \times 100 \\ = 3 \times 8.32 \times 50 J \end{array}$

$Δ U = 1 248J = 1. 25 \times 1 0^{3} J$

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One mole of an ideal monoatomic gas is heated at a constant pressure from 0∘C to 100∘C. Then the change in the internal energy of the gas is (Given, R=8.32Jmol−1K−1 )

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One mole of an ideal monoatomic gas is heated at a constant pressure from $0^{\circ} C$ to $100^{\circ} C .$ Then the change in the internal energy of the gas is (Given, $R = 8.32 J m o l^{- 1} K^{- 1}$ )