5.6, L of helium gas at STP is adiabatically | vedclass

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(A) For an ideal gas,the number of moles $n$ at $STP$ is given by $n = \frac{V}{22.4} = \frac{5.6}{22.4} = 0.25 = \frac{1}{4} \, mol$.Helium is a mono

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$\frac{9}{8} R T_1$

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(A) For an ideal gas,the number of moles $n$ at $STP$ is given by $n = \frac{V}{22.4} = \frac{5.6}{22.4} = 0.25 = \frac{1}{4} \, mol$.Helium is a monoatomic gas,so the adiabatic index $\gamma = \frac{5}{3}$.For an adiabatic process,the relation between temperature and volume is $T_1 V_1^{\gamma-1} = T_2 V_2^{\gamma-1}$.Substituting the values: $T_2 = T_1 \left( \frac{V_1}{V_2} ight)^{\gamma-1} = T_1 \left( \frac{5.6}{0.7} ight)^{\frac{5}{3}-1} = T_1 (8)^{2/3} = T_1 (2^3)^{2/3} = 4 T_1$.The work done in an adiabatic process is given by $W = \frac{nR(T_1 - T_2)}{\gamma - 1}$.Since it is compression,the magnitude of work done is $|W| = \frac{nR(T_2 - T_1)}{\gamma - 1}$.Substituting the values: $|W| = \frac{(1/4) R (4 T_1 - T_1)}{(5/3) - 1} = \frac{(1/4) R (3 T_1)}{2/3} = \frac{3/4 R T_1}{2/3} = \frac{3}{4} 	imes \frac{3}{2} R T_1 = \frac{9}{8} R T_1$.

$5.6\, L$ of helium gas at $STP$ is adiabatically compressed to $0.7\, L$. Taking the initial temperature to be $T_1$,the magnitude of work done in the process is

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