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

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

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, then room will become hot, because amount of heat removed would be less than the amount of heat released in the room.

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.

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.