# Heat relation of a isobaric expansion

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.

• 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 – LSS Sep 10 at 15:41
• No, integration doesn't have any sense but I didn't understand the meaning of "a factor $x$". Thank you for the answer. – psidaga Sep 10 at 15:55
• Why does the last relation have a term $\Delta U$? In any case your way was undoubtedly more direct. – psidaga Sep 10 at 15:58