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A gas takes part in two processes in which it is heated from the same initial state $1$ to the same final temperature. The processes are shown on the $P-V$ diagram by the straight line $1-2$ and $1-3$. $2$ and $3$ are the points on the same isothermal curve. $Q_1$ and $Q_2$ are the heat transfer along the two processes. Then
$Q_1 = Q_2$
$Q_1 < Q_2$
$Q_1 > Q_2$
insufficient data
Solution
Let the volumes and temperatures of gas at points 1,2 and 3 are $V_1, T_1, V_2, T_2$ and $V _3, T _3$ respectively.
Given: $T_2=T_3$
Thus $\Delta U _1= nC _{ v }\left( T _2- T _1\right)$
Also $\Delta U _2= nC _{ v }\left( T _3- T _1\right)$
But $T _2= T _3 \Rightarrow \Delta U _1=\Delta U _2$
Work done by the gas in process 1, $W _1= nRT \ln \frac{ V _2}{ V _1}$
Also work done by gas in process $2, \quad W_2=n R T \ln \frac{V_3}{V_1}$
$V _3 > V _2 \Rightarrow W _2 > W _1$
From ist law: $\quad Q=\Delta U+W$
$\therefore \quad Q _{ I }=\Delta U _1+ W _1$
Also $\quad Q _2=\Delta U _2+ W _2$
$\therefore Q _2- Q _1= W _2- W _1>0 \quad\left(\because \Delta U _1=\Delta U _2\right)$
$\Rightarrow Q _2 > Q _1$
Similar Questions
One mole of a monatomic ideal gas is taken through a cycle $ABCDA$ as shown in the $P-V$ diagram. Column $II$ gives the characteristics involved in the cycle. Match them with each of the processes qiven in Column $I$
| Column $I$ | Column $II$ |
| $(A)$ Process $A \rightarrow B$ | $(p)$ Internal energy decreases. |
| $(B)$ Process $B \rightarrow C$ | $(q)$ Internal energy increases. |
| $(C)$ Process $C \rightarrow D$ | $(r)$ Heat is lost. |
| $(D)$ Process $D \rightarrow A$ | $(s)$ Heat is gained. |
| $(t)$ Work is done on the gas. |
