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9
MCQ (Multiple Correct Answer)

Figure shows the $p-V$ diagram of an ideal gas undergoing a change of state from $A$ to $B$. Four different parts I, II, III and IV as shown in the figure may lead to the same change of state.

A
Change in internal energy is same in IV and III cases, but not in I and II
B
Change in internal energy is same in all the four cases
C
Work done is maximum in case I
D
Work done is minimum in case II
10
MCQ (Multiple Correct Answer)

Consider a cycle followed by an engine (figure.)

1 to 2 is isothermal

2 to 3 is adiabatic

3 to 1 is adiabatic

Such a process does not exist, because

A
heat is completely converted to mechanical energy in such a process, which is not possible
B
mechanical energy is completely converted to heat in this process, which is not possible
C
curves representing two adiabatic processes don't intersect
D
curves representing an adiabatic process and an isothermal process don't intersect
11
MCQ (Multiple Correct Answer)

Consider a heat engine as shown in figure. $Q_1$ and $Q_2$ are heat added both to $T_1$ and heat taken from $T_2$ in one cycle of engine. $W$ is the mechanical work done on the engine.

If W > 0, then possibilities are

A
$Q_1>Q_2>0$
B
$Q_2>Q_1>0$
C
$Q_2< Q_1<0$
D
$Q_1<0, Q_2>0$
12
Subjective

Can a system be heated and its temperature remains constant?

Explanation

Yes, this is possible when the entire heat supplied to the system is utilised in expansion. i.e., its working against the surroundings.

13
Subjective

A system goes from $P$ to $Q$ by two different paths in the $p-V$ diagram as shown in figure. Heat given to the system in path 1 is 1000 J . The work done by the system along path 1 is more than path 2 by 100 J . What is the heat exchanged by the system in path 2?

Explanation

$$ \begin{aligned} \text { For path 1, } \quad \text { Heat given } Q_1 & =+1000 \mathrm{~J} \\ \text { Work done } & =W_1 \text { (let) } \end{aligned}$$

For path 2,

Work done $$\left(W_2\right)=\left(W_1-100\right) \mathrm{J}$$

Heat given $Q_2=$ ?

$$\begin{aligned} &\text { As change in internal energy between two states for different path is same. }\\ &\begin{aligned} \therefore \quad & \Delta U =Q_1-W_1=Q_2-W_2 \\ \Rightarrow \quad & 1000-W_1 =Q_2-\left(W_1-100\right) \\ \Rightarrow \quad & Q_2 =1000-100=900 \mathrm{~J} \end{aligned} \end{aligned}$$