The vapour pressure of benzene is 1-53- 10 4 Nm -2 at 303K and 5-2- 10 4 Nm -2 at 333K- The latent heat of evaporation kJ of benzene over this temperature range is-

2.303log( P _ 2 / P _ 1 )=Δ H _ v / R ( T _ 2 T _ 1 / T _ 1 T _ 2 ) #160;2.303 log ( #160; 5.2× 10 ^ 4 / 1.53× 10 ^ 4 #160; )=Δ H _ v / 8.31 ...

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Standard XII

Chemistry

Thermodynamic Processes

The vapour pr...

Question

The vapour pressure of benzene is $1.53 \times 10^{4} {N m}^{- 2}$ at $303 K$ and $5.2 \times 10^{4} {N m}^{- 2}$ at $333 K$ . The latent heat of evaporation $(k J)$ of benzene over this temperature range is:

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Solution

$2.303 log (\frac{P_{2}}{P_{1}}) = \frac{{Δ H}_{v}}{R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})$
$2.303 l o g (\frac{5.2 \times 10^{4}}{1.53 \times 10^{4}}) = \frac{{Δ H}_{v}}{8.314} (\frac{333 - 303}{333 \times 303})$

$∴ {Δ H}_{v} = 34215 J = 34.215 k J$

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The vapour pressure of benzene is 1.53×104Nm−2 at 303K and 5.2×104Nm−2 at 333K. The latent heat of evaporation(kJ) of benzene over this temperature range is:

2.303log(P2P1)=ΔHvR(T2−T1T1T2)2.303log(5.2×1041.53×104)=ΔHv8.314(333−303333×303)∴ ΔHv=34215J=34.215 kJ

The vapour pressure of benzene is $1.53 \times 10^{4} {N m}^{- 2}$ at $303 K$ and $5.2 \times 10^{4} {N m}^{- 2}$ at $333 K$ . The latent heat of evaporation $(k J)$ of benzene over this temperature range is:

$2.303 log (\frac{P_{2}}{P_{1}}) = \frac{{Δ H}_{v}}{R} (\frac{T_{2} - T_{1}}{T_{1} T_{2}})$
$2.303 l o g (\frac{5.2 \times 10^{4}}{1.53 \times 10^{4}}) = \frac{{Δ H}_{v}}{8.314} (\frac{333 - 303}{333 \times 303})$

$∴ {Δ H}_{v} = 34215 J = 34.215 k J$