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Exercises · 5.2

Q.A vessel of 120 mL capacity contains a certain amount of gas at 35 °C and 1.2 bar pressure. The gas is transferred to another vessel of volume 180 mL at 35 °C. What would be its pressure?

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✓ Free question

Step 1 – Identify the process

Same gas, same temperature (35°C) in both vessels, fixed amount of gas — Boyle's law:

p1V1=p2V2p_1V_1 = p_2V_2

Step 2 – Data

p1=1.2 bar,V1=120 mL,V2=180 mLp_1 = 1.2\ \text{bar}, \quad V_1 = 120\ \text{mL}, \quad V_2 = 180\ \text{mL}

Step 3 – Solve

p2=p1V1V2=(1.2 bar)(120 mL)180 mL=144180 bar=0.8 barp_2 = \frac{p_1V_1}{V_2} = \frac{(1.2\ \text{bar})(120\ \text{mL})}{180\ \text{mL}} = \frac{144}{180}\ \text{bar} = 0.8\ \text{bar}

✓Final answer

p2=0.8 bar\boxed{p_2 = 0.8\ \text{bar}}

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