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?
$$ \begin{aligned} \text { For path 1, } \quad \text { Heat given } Q_1 & =+1000 \mathrm{~J} \\ \text { Work done } & =W_1 \text { (let) } \end{aligned}$$
For path 2,
Work done $$\left(W_2\right)=\left(W_1-100\right) \mathrm{J}$$
Heat given $Q_2=$ ?
$$\begin{aligned} &\text { As change in internal energy between two states for different path is same. }\\ &\begin{aligned} \therefore \quad & \Delta U =Q_1-W_1=Q_2-W_2 \\ \Rightarrow \quad & 1000-W_1 =Q_2-\left(W_1-100\right) \\ \Rightarrow \quad & Q_2 =1000-100=900 \mathrm{~J} \end{aligned} \end{aligned}$$
If a refrigerator's door is kept open, will the room become cool or hot? Explain.
If a refrigerator's door is kept open, then room will become hot, because amount of heat removed would be less than the amount of heat released in the room.
Is it possible to increase the temperature of a gas without adding heat to it? Explain.
Yes, during adiabatic compression the temperature of a gas increases while no heat is given to it.
$$\begin{aligned} \text{In adiabatic compression,}\quad & d Q=0 \\ \therefore \quad \text{From first law of thermodynamics,}\quad & d U=d Q-d W \\ & d U=-d W \end{aligned}$$
In compression work is done on the gas i.e., work done is negative.
Therefore, $$\quad d U=\text { Positive }$$
Hence, internal energy of the gas increases due to which its temperature increases.
Air pressure in a car tyre increases during driving. Explain.
During, driving, temperature of the gas increases while its volume remains constant.
So, according to Charle's law, at constant volume ($V$),
Pressure $(p) \propto$ Temperature $(T)$
Therefore, pressure of gas increases.
Consider a Carnot's cycle operating between $T_1=500 \mathrm{~K}$ and $T_2=300 \mathrm{~K}$ producing 1 kJ of mechanical work per cycle. Find the heat transferred to the engine by the reservoirs.
$$\begin{aligned} \text{Given,}\quad \text { temperature of the source } T_1 & =500 \mathrm{~K} \\ \text { Temperature of the sink } T_2 & =300 \mathrm{~K} \\ \text { Work done per cycle } W & =1 \mathrm{~kJ}=1000 \mathrm{~J} \end{aligned}$$
Heat transferred to the engine per cycle $Q_1=$ ?
Efficiency of a Carnot engine $(\eta)=1-\frac{T_2}{T_1}=1-\frac{300}{500}=\frac{200}{500}=\frac{2}{5}$
$$\text{and}\quad \eta=\frac{W}{Q_1}$$
$$\Rightarrow \quad Q_1=\frac{W}{\eta}=\frac{1000}{(2 / 5)}=2500 \mathrm{~J}$$