Question

One mole of an ideal gas is taken from state $A$ to state $B$ by three different processes $\left(a\right)ACB$, $\left(b\right)ADB$ and $\left(c\right)AEB$ as shown in the $P-V$ diagram. The heat absorbed by the gas is

• A. Greater in process $\left(b\right)$ than in $\left(a\right)$
• B. The least in process $\left(b\right)$
• C. The same in $\left(a\right)$ and $\left(c\right)$
• D. Less in $\left(c\right)$ than in $\left(b\right)$

Answer 1

The correct option is D Less in $\left(c\right)$ than in $\left(b\right)$

From first law of thermodynamics Heat absorbed by gas in three processes is given by,
$Q_{ACB}=\Delta U+W_{ACB}...\left(i\right)$
$Q_{ADB}=\Delta U...\left(ii\right)$
$(\because W_{ADB}=0,\text{as}V=\text{constant})$
$Q_{AEB}=\Delta U+W_{AEB}...\left(iii\right)$
The change in internal energy in all the three cases is same i.e $\Delta U$ for each process.

$\Rightarrow W_{ACB}$ is $+ve$ because area under curve during expansion $(C\to B)$ is greater than area under curve during compression$(A\to C)$

$\Rightarrow W_{AEB}$ is $-ve$ because area under curve during expansion $(A\to E)$ is less than area under curve during compression$(E\to B)$

Hence on comparing $\left(i\right),\left(ii\right),\left(iii\right)$ we get,
$\Rightarrow Q_{ACB}>Q_{ADB}>Q_{AEB}$