Calculate q- for the reversible isothermal expansion of one mole of an ideal gas at 127- C from a volume of 10 dm 3 to 20 dm 3 Take ln 2-0-3 and R -8-3 JK -1 mol -1A- 2480-78 JB- -2293-78 JC- -2480-78 JD- 2293-78 J

The correct option is D 2293.78 JSince the process is isothermal nbsp; E = H = 0 according to first law of thermodynamics, E = q nbsp; + w = 0 q = W Ag ...

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Standard XII

Chemistry

Work Done in Isothermal Reversible Process

Calculate q, ...

Question

Calculate q, for the reversible isothermal expansion of one mole of an ideal gas at $127^{\circ} C$ from a volume of 10 dm $^{3}$ to 20 dm $^{3}$ Take ln2 = 0.3 and R = 8.3 $J K^{- 1} m o l^{- 1}$

2293.78 J

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2480.78 J

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-2293.78 J

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-2480.78 J

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Solution

The correct option is A 2293.78 J
Since the process is isothermal
$△ E = △ H = 0$
according to first law of thermodynamics,
$△ E = q + w = 0 q = - W$
Again, for isothermal process,
$W = - 2.303 n R T log \frac{V_{2}}{V_{1}} = - 2.303 \times 1 \times 8.3 \times 400 log \frac{20}{10} = - 2.303 \times 1 \times 8.3 \times 400 \times 0.3 J = - 2293.78 J$
Again, q = - W
= 2293.78 J

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Calculate q, for the reversible isothermal expansion of one mole of an ideal gas at 127∘C from a volume of 10 dm3 to 20 dm3 Take ln2 = 0.3 and R = 8.3 JK−1mol−1

The correct option is A 2293.78 JSince the process is isothermal △E=△H=0 according to first law of thermodynamics, △E=q+w=0q=−W Again, for isothermal process, W=−2.303nRTlogV2V1=−2.303×1×8.3×400log2010=−2.303×1×8.3×400×0.3 J=−2293.78 J Again, q = - W = 2293.78 J

Calculate q, for the reversible isothermal expansion of one mole of an ideal gas at $127^{\circ} C$ from a volume of 10 dm $^{3}$ to 20 dm $^{3}$ Take ln2 = 0.3 and R = 8.3 $J K^{- 1} m o l^{- 1}$