5.6 L of helium gas at STP is adiabatically | vedclass

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(B) For an adiabatic process,the work done is given by $W = \frac{nR(T_i - T_f)}{\gamma - 1}$.Helium is a monoatomic gas,so $\gamma = 5/3$.At $STP$,$1

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$-\left(\frac{9}{8} RT\right)$

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(B) For an adiabatic process,the work done is given by $W = \frac{nR(T_i - T_f)}{\gamma - 1}$.Helium is a monoatomic gas,so $\gamma = 5/3$.At $STP$,$1 \ mole$ of gas occupies $22.4 \ L$. Given $V_i = 5.6 \ L$,the number of moles $n = \frac{5.6}{22.4} = 0.25 \ mole = \frac{1}{4} \ mole$.For an adiabatic process,$T_i V_i^{\gamma-1} = T_f V_f^{\gamma-1}$.$T_f = T_i \left(\frac{V_i}{V_f}\right)^{\gamma-1} = T \left(\frac{5.6}{0.7}\right)^{(5/3)-1} = T(8)^{2/3} = T(2^3)^{2/3} = 4T$.Now,$W = \frac{n R (T - 4T)}{(5/3) - 1} = \frac{(1/4) R (-3T)}{2/3} = \frac{-3/4 RT}{2/3} = -\frac{9}{8} RT$.Since the gas is compressed,work is done on the gas,so the work done by the gas is negative.

$5.6 \ L$ of helium gas at $STP$ is adiabatically compressed to $0.7 \ L$. If the initial temperature of the gas is $T \ K$,the work done in the process is ($R$ is the universal gas constant in $SI$ units).

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