An ideal monoatomic gas expands at constant pressure. The work done by the gas on its environment is $200…
An ideal monoatomic gas expands at constant pressure. The work done by the gas on its environment is $200 \mathrm{~J}$, then the heat absorbed by the gas during the process is
$500 \mathrm{~J}$
$300 \mathrm{~J}$
$200 \mathrm{~J}$
$600 \mathrm{~J}$
Solution
Given, $W=200 \mathrm{~J}$
Since, ideal monoatomic gas expands at constant pressure, hence heat absorbed, $Q=n C_p \Delta T$
Work done, $W=n R \Delta T$
$
\begin{aligned}
& \frac{Q}{W}=\frac{n C_p \Delta T}{n R \Delta T} \\
& \frac{Q}{W}=\frac{C_p}{R}
...(i)\end{aligned}
$
for monoatomic gas, $C_V=1.5 R$
$
\begin{aligned}
C_p & =C_v+R \\
& =1.5 R+R
\end{aligned}
$
$
\Rightarrow \quad C_p=2.5 R
$
From Eqs (i) and (ii), we get
$
\begin{aligned}
\frac{Q}{W} & =\frac{2.5 R}{R} \\
& =2.5 \\
Q & =2.5 \mathrm{~W} \\
& =25 \times 200 \\
& =500 \mathrm{~J}
\end{aligned}
$