Question

Two moles of an ideal gas at temperature $T_0=300K$ was cooled isochorically so that the pressure was reduced to half. Then, in an isobaric process, the gas expanded till its temperature got back to the initial value. Find the total amount of heat absorbed(in J) by the gas in the process.

Answer 1

In an isochoric process, $\Delta W=0$ $(\Delta V=0)$
and $\Delta U=C_v\Delta T=\frac{R}{\gamma -1}\Delta T$ $(\because C_v=\frac{R}{\gamma -1})$
$⟹\Delta U=\frac{\Delta \left(RT\right)}{\gamma -1}=\frac{\Delta \left(pV\right)}{\gamma -1}$
$=\frac{V}{\gamma -1}\Delta p=\frac{V}{\gamma -1}(\frac{1}{2}p-p)=-\frac{1}{2}\frac{pV}{\gamma -1}$
Now $\Delta Q=\Delta U+\Delta W$ (always)
$\therefore \Delta Q=-\frac{1}{2}\frac{pV}{\gamma -1}+0=-\frac{1}{2}\frac{pV}{\gamma -1}=-\frac{1}{2\gamma }\frac{R{T}_0}{-1}$
The negative sign shows that heat is not added but subtracted from the gas in the process.
In the isobaric process
$\Delta W={\int }_{V_i}^{V_f}\frac{p}{2}dv=\frac{p}{2}(V_f-V_i)\left[Inthisprocesspressureis\frac{P}{2}\right]$
In an isochoric process, $V$ remains constant, so $p\alpha T$ and temperature is reduced to $\frac{T_0}{2}$.
In the isobaric process, the original temperature is restored.
$\Delta T=T_0-\frac{T_0}{2}=\frac{T_0}{2}$
$\Delta U=C_v\Delta T=\frac{R}{\gamma -1}\frac{T_0}{2}$ $(\because C_v=\frac{R}{(\gamma -1)}and\Delta T=\frac{T_0}{2})$
$\therefore \Delta Q=\Delta U+\Delta W=\frac{1}{2}\frac{R{T}_0}{\gamma -1}+\frac{1}{2}R{T}_0$
$\therefore$ Net heat added is
$[-\frac{1}{2}\frac{R{T}_0}{\gamma -1}]+[\frac{1}{2}\frac{R{T}_0}{\gamma -1}+\frac{1}{2}R{T}_0]=\frac{1}{2}R{T}_0$
This is the heat added when the number of moles is one. When there are $2moles$, heated is
$2\times \frac{1}{2}R{T}_0=R{T}_0=8.3\times 300=2490J$