An ideal gas expanded irreversibly against 10 bar pressure from $20 \mathrm{~L}$ to $30 \mathrm{~L}$.…
An ideal gas expanded irreversibly against 10 bar pressure from $20 \mathrm{~L}$ to $30 \mathrm{~L}$. Calculate $Q$ if the process is isoenthalpic. $1 \mathrm{~L} \mathrm{bar}=100 \mathrm{~J}$
0
$100 \mathrm{~J}$
$-100 \mathrm{~J}$
$10 \mathrm{~kJ}$
Solution
According to thermodynamics irreversibly expanded system work.
$
W_{\text {irr }}=-p_{\text {opp. }}\left(V_2-V_1\right)...(i)
$
Ideal gas expanded irreversibly $=10 \mathrm{bar}$
$
\begin{aligned}
& V_2=30 \mathrm{~L} \\
& V_1=20 \mathrm{~L}
\end{aligned}
$
Put all value in Eq. (i), we get
$
\begin{aligned}
W_{\text {irr }} & =-10 \text { bar }(30-20) \mathrm{L} \\
& =-9.869 \mathrm{~atm} \times 10 \mathrm{~L}
\end{aligned}
$
$
\begin{aligned}
& =-9999.76 \mathrm{~J} \\
& =-10 \mathrm{~kJ}
\end{aligned}
$
Since, for expanded irreversibly, $\Delta H=0$
$\therefore(\Delta H=0$ i.e. isoenthalpic process)
$
\begin{aligned}
& Q=-W_{\text {irr }} \\
& Q=-(-10 \mathrm{~kJ}) \\
& Q=10 \mathrm{~kJ}
\end{aligned}
$
Hence, $Q$ value of isoenthalpic process is $10 \mathrm{~kJ}$