A piston closed at the top contains $n=0,2$ $mol$ of a mono-atomic ideal gas at the temperature of $T_0$. The pressure of the gas is constant during the transformation and we don't have to consider the heat capacity of the piston(we have a isobaric expansion at the pressure $p_0$).

Now we tranfer a quantity of heat $Q$ to the piston and the gas starts expanding its volume. The quantity of heat $Q$ is proportional to $\Gamma dV$, where $\Gamma$ is a constant. I have to find the value of the constant $\Gamma$.

My idea is this: the gas is ideal so we can consider the relation $p_0V_0=nRT_0 \implies V_0=\dfrac{nRT_0}{p_0}$, let's consider $p_0=p_{atm}$.
The pressure is constant during the expansion, so $p_0=\dfrac{nRT_0}{V_0}=\dfrac{nRT_1}{V_1}=p_1\implies T_1=\dfrac{V_1}{V_0} T_0$.

Now if we calculate $Q$ of the isobaric expansion: $$Q=nc_p\Delta T=\dfrac{5}{2}nR(T_1-T_0)=\dfrac{5}{2}nRT_0\Big(\dfrac{V_1}{V_0}-1\Big)\\=\dfrac{5}{2}nRT_0\Bigg(\dfrac{V_1-V_0}{\dfrac {nRT_0}{p_0}}\Bigg)=\dfrac{5}{2}p_0(V_1-V_0)$$ So, in conclusion we get $\Gamma=\dfrac{5}{2}p_0=\dfrac{5}{2}p_{atm}$ if we consider the difference of the volume infinitesimal.

Thank you in advance.

  • 1
    $\begingroup$ That is not really necessary, and i think doesn't make sense integrate the heat. $$ U = nCRT $$ $$dU = dQ - dW = dQ - PdV $$ $$dQ = CnRdT + PdV = \frac{CnRT_{o}}{V_{o}}dV + PdV = (\frac{CnRT_{o}}{V_{o}} + P)dV = (CP + P)dV = P(C+1)dv $$ $$ \Delta Q = P(C+1)\Delta U$$ So do your know C, and know P $\endgroup$ – LSS Sep 10 at 15:41
  • $\begingroup$ No, integration doesn't have any sense but I didn't understand the meaning of "a factor $x$". Thank you for the answer. $\endgroup$ – psidaga Sep 10 at 15:55
  • $\begingroup$ Why does the last relation have a term $\Delta U$? In any case your way was undoubtedly more direct. $\endgroup$ – psidaga Sep 10 at 15:58
  • $\begingroup$ Ops, error of digitation. $\endgroup$ – LSS Sep 10 at 21:46
  • $\begingroup$ I am trying to edit but., i think is not more possible, it would be Delta V. I am not sure if i get... factor x? $\endgroup$ – LSS Sep 10 at 21:47

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