Question #74b54

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Question #74b54

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Answer 1 · 2017-03-28T04:50:59

The new volume of the ethane gas is 1.87 L.

Explanation:

In order to solve this question, we need to use Charles's law which states that the volume of a gas is proportional to its temperature in Kelvin. Mathematically,

$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$ $V_1/T_1 = V_2/T_2$

Where $V_{1}$ $V_1$ is the initial volume of the gas, $T_{1}$ $T_1$ is the initial temperature of the gas (in Kelvin), $V_{2}$ $V_2$ is the volume of the gas after heating (or cooling, if that's what you did), and $T_{2}$ $T_2$ is the final temperature of the gas (in Kelvin).

Note: Please remember that the temperature is in Kelvin scale not in Degree scale.

Now, from the question we have,

Initial Volume of Ethane gas ( $V_{1}$ $V_1$ ) = 4.17 L
Initial Temperature of Ethane gas ( $T_{1}$ $T_1$ ) = 725 + 273 = 998 K
Final Volume of Ethane gas ( $V_{2}$ $V_2$ ) = ?
Final Temperature of Ethane gas ( $T_{2}$ $T_2$ ) = 175 + 273 = 448 K

Using the Charle's law:

$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$ $V_1/T_1 = V_2/T_2$

Plug the given values in above laws, we get:

or, $\frac{4.17 L}{998 K} = \frac{V_{2}}{448 K}$ ${4.17 L}/{998 K} = V_2/{448 K}$

Multiplying both sides by 448 K, we get:

or, $\frac{4.17 L}{998 K} X 448 K = V_{2}$ ${4.17 L}/{998 K}X448 K = V_2$

Rearranging,

or, $V_{2} = \frac{4.17 L}{998 K} X 448 K$ $V_2={4.17 L}/{998 K}X448 K$

or, $V_{2} = 1.87 L$ $V_2=1.87 L$

Therefore, the new volume is 1.87 L .

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Question #74b54

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