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Home  >>  JEEMAIN and AIPMT  >>  Physics  >>  Class11  >>  Thermodynamics
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A sample of Ar gas at 1 atm pressure and $27^{\circ}C$ expands reversibly and adiabatically from $1.25\: dm^3\: to\: 2.5\: dm^3$. Calculate enthalpy change of this process. $CV$ for argon is $ 12.4\: kJ/mol$

$\begin {array} {1 1} (A)\;1145.7 \: J & \quad (B)\;-1145.7 \: J \\ (C)\;114.57 \: J & \quad (D)\;-114.57 \: J \end {array}$

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For adiabatic process,$ \Delta H = nC_p. \Delta T$
Now, $C_p-C_v = R \; or\: , C_p = R+C_v = 8.314 + 12.48 = 20.794 \: J/Kmol$
Also, $ \gamma-1 = \large\frac{C_p}{C_v}$$ – 1 = \large\frac{(C_p-C_v)}{C_v}$$ = \large\frac{R}{C_v}$$ = \large\frac{8.314}{12.48}$$ = 0.66$
$\large\frac{T_2}{T_1}$$ = \bigg( \large\frac{V_1}{V_2} \bigg)^{\gamma-1}$ or, $ T_2 = T_1 \bigg(\large\frac{V_1}{V_2} \bigg) ^{0.66}$$ = 300 \times \bigg(\large\frac{125}{250} \bigg)^{0.66}$
or,$T_2 = 189.8\: K$
So,$ \Delta T = T_2 – T_1 = (189.8 – 300) = -110.2$
$PV = nRT$
$\Rightarrow n = \large\frac{PV}{RT} $$= \large\frac{1 \times 1.25}{(0.082 \times 300) }$$= 0.05$
$\Delta H = nC_p \Delta T = 0.05 \times 20.794 \times (-110.2) = -114.57 \: J$
Ans : (B)
answered Mar 20, 2014 by thanvigandhi_1
 

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