On a ship sailing in a Pacific ocean where the temperature is ${23.4}^{o} C$ , 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 ${26.1}^{o} C$ ?
(Write the answer to the nearest integer)

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Solution

According to Charles' law
$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}; V_{1} = 2 L; V_{2} = ?$
$T_{1} = 273 + 23.4 = 296.4 K$
$T_{2} = 273 + 26.1 = 299.1$
$V_{2} = \frac{V_{1} T_{2}}{T_{1}}$
$= \frac{2 L \times 299.1 K}{296.4 K} = 2.018 \approx 2 L$

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On a ship sailing in a Pacific ocean where the temperature is 23.4oC, 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 26.1oC?(Write the answer to the nearest integer)

According to Charles' lawV1T1=V2T2;V1=2L;V2=?T1=273+23.4=296.4KT2=273+26.1=299.1V2=V1T2T1=2L×299.1K296.4K=2.018≈2L

On a ship sailing in a Pacific ocean where the temperature is ${23.4}^{o} C$ , 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 ${26.1}^{o} C$ ?
(Write the answer to the nearest integer)

According to Charles' law
$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}; V_{1} = 2 L; V_{2} = ?$
$T_{1} = 273 + 23.4 = 296.4 K$
$T_{2} = 273 + 26.1 = 299.1$
$V_{2} = \frac{V_{1} T_{2}}{T_{1}}$
$= \frac{2 L \times 299.1 K}{296.4 K} = 2.018 \approx 2 L$