$A$ vessel of $120 \, mL$ capacity contains a certain amount of gas at $35^{\circ} C$ and $1.2 \, bar$ pressure. The gas is transferred to another vessel of volume $180 \, mL$ at $35^{\circ} C$. What would be its pressure?

(0.8 BAR) Given:
Initial pressure,$p_{1} = 1.2 \, bar$
Initial volume,$V_{1} = 120 \, mL$
Final volume,$V_{2} = 180 \, mL$
Since the temperature remains constant,the final pressure $(p_{2})$ can be calculated using Boyle's law.
According to Boyle's law,$p_{1} V_{1} = p_{2} V_{2}$.
$p_{2} = \frac{p_{1} V_{1}}{V_{2}}$
$p_{2} = \frac{1.2 \times 120}{180} \, bar$
$p_{2} = 0.8 \, bar$
Therefore,the final pressure of the gas would be $0.8 \, bar$.