Question

initially an ideal diatomic gas pressure $p_i = 2.5\times10^{5} pa$ and volume $v_i = 4.7\times10^{-4} m^{3}$. how much work does it do and what is the change in its internal energy if it expands adiabatically to final volume of $v_f = 8.2\times10^{-4} m^{3}$? throughout the process molecule have rotate but not oscillate.
$t = t_{trans.}+rot.$
$f = 3 + 2=5$
$phi = 0, pv^{delta}=const.$
$phi=delta e+w$
$delta e=-w =-int pdv$
$pv = nrt$
$p=\frac{nrt}{v}$
$delta=\frac{c_p}{c_v}=\frac{c_v + r}{c_v}=\frac{\frac{f}{2}r+r}{\frac{f}{2}r}=\frac{\frac{5}{2}r + r}{\frac{5}{2}r}=\frac{7}{5}$
$delta e=-\frac{p_iv_i^{delta}}{ 1-delta}v_f^{-delta + 1}-v_i^{-delta + 1}$
$w (q = 0)$
$delta e=-(2.5\times10^{5}\times4.7\times10^{-4})\times\frac{1}{1 - \frac{7}{5}}\times(8.2\times10^{-4})^{-\frac{7}{5}+1}-(4.7\times10^{-4})^{-\frac{7}{5}+1}$
$=-? j$

Explanation:

Step1: Identify adiabatic work formula

For an adiabatic process of an ideal gas, the work done $W = \frac{P_iV_i - P_fV_f}{\gamma - 1}$, and also $\Delta E=-W$. Given $P_i = 2.5\times10^{5}\ Pa$, $V_i = 4.7\times10^{-6}\ m^{3}$, $V_f=8.2\times10^{-6}\ m^{3}$, and for a di - atomic gas with only rotation (no oscillation) $\gamma=\frac{C_p}{C_v}=\frac{\frac{f}{2}R + R}{\frac{f}{2}R}$, where $f = 5$ (3 translational + 2 rotational), so $\gamma=\frac{7}{5}$.

Step2: Calculate work done

First, use the adiabatic relation $P_iV_i^{\gamma}=P_fV_f^{\gamma}$, so $P_f = P_i(\frac{V_i}{V_f})^{\gamma}$. Then the work done $W=\frac{P_iV_i - P_i(\frac{V_i}{V_f})^{\gamma}V_f}{\gamma - 1}$. Substitute the values:

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Since $\Delta E=-W$, $\Delta E=- 1.015\ J$

Answer:

Work done $W = 1.015\ J$, Change in internal energy $\Delta E=-1.015\ J$