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Standard XII

Chemistry

Pressure and Temperature

The fall in t...

Question

The fall in temperature of helium gas initially at 20C when it is suddenly expanded to 8 times its original volume is: $(γ = \frac{5}{3})$

70.25 K

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71.25 K

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72.25 K

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73.25 K

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Solution

The correct option is A 73.25 K
Expanded suddenly $\to$ Adiabatic process
${T V}^{Y - 1} =$ constant
${T V}^{5 / 3 - 1} =$ const
${T V}^{2 / 3} =$ const
$T_{1} V_{1}^{2 / 3} = T_{2} V_{2}^{2 / 3}$

$\Rightarrow \frac{T_{1}}{T_{2}} = {(\frac{V_{2}}{V_{1}})}^{2 / 3}$

$\Rightarrow \frac{T_{1}}{T_{2}} = {(8)}^{2 / 3}$

$\Rightarrow \frac{(20 + 273) K}{T_{2}} = 4$

$\Rightarrow T_{2} = \frac{293}{4} K$

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The fall in temperature of helium gas initially at 20C when it is suddenly expanded to 8 times its original volume is: (γ=53)

The correct option is A 73.25 KExpanded suddenly → Adiabatic processTVY−1= constantTV5/3−1= constTV2/3= constT1V2/31=T2V2/32⇒T1T2=(V2V1)2/3⇒T1T2=(8)2/3⇒(20+273)KT2=4⇒T2=2934K

The fall in temperature of helium gas initially at 20C when it is suddenly expanded to 8 times its original volume is: $(γ = \frac{5}{3})$