On a ship sailing in the Pacific ocean, where the temperature is 27 $^{\circ}$ C, a balloon is filled with 2 L air. What will be the volume of the balloon when the ship reaches the Indian ocean where the temperature is 29 $^{\circ}$ C ? (Assume pressure is constant) [2 marks]

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Solution

According to Charles' law:

$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$ [0.5 marks]

$V_{1}$ = 2 L, $V_{2}$ = ?

$T_{1}$ = 273 + 27 = 300K [0.5 marks]
$T_{2}$ = 273 + 29 = 3021K [0.5 marks]

Therefore, $V_{2} = \frac{V_{1} \times T_{2}}{T_{1}}$ = $\frac{2 \times 302}{300}$ = 2.01 L [0.5 marks]

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On a ship sailing in the Pacific ocean, where the temperature is 27 ∘C, a balloon is filled with 2 L air. What will be the volume of the balloon when the ship reaches the Indian ocean where the temperature is 29 ∘C ? (Assume pressure is constant) [2 marks]

According to Charles' law: V1T1=V2T2 [0.5 marks] V1 = 2 L, V2 = ? T1 = 273 + 27 = 300K [0.5 marks] T2 = 273 + 29 = 3021K [0.5 marks] Therefore, V2=V1×T2T1= 2×302300 = 2.01 L [0.5 marks]

On a ship sailing in the Pacific ocean, where the temperature is 27 $^{\circ}$ C, a balloon is filled with 2 L air. What will be the volume of the balloon when the ship reaches the Indian ocean where the temperature is 29 $^{\circ}$ C ? (Assume pressure is constant) [2 marks]

According to Charles' law:

$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$ [0.5 marks]

$V_{1}$ = 2 L, $V_{2}$ = ?

$T_{1}$ = 273 + 27 = 300K [0.5 marks]
$T_{2}$ = 273 + 29 = 3021K [0.5 marks]

Therefore, $V_{2} = \frac{V_{1} \times T_{2}}{T_{1}}$ = $\frac{2 \times 302}{300}$ = 2.01 L [0.5 marks]