Question #5bfc7

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Question #5bfc7

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Answer 1 · 2016-06-05T02:05:30

The new volume is $7.1 L$ $"7.1 L"$

Explanation:

Let's start off with identifying our known and unknown variables.
The first volume we have is $6.0 L$ $"6.0 L"$ , the first temperature is ${8.0}^{\circ} C$ $8.0^@"C"$ , and the second temperature is ${60.0}^{\circ} C$ $60.0^@"C"$ . Our only unknown is the second volume.

We can obtain the answer using Charles' Law which shows that there is a direct relationship between volume and temperature as long as the pressure and number of moles remain unchanged.

The equation we use is

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

where the numbers $1$ $1$ and $2$ $2$ represent the first and second conditions. I must also add that the volume must have units of liters and the temperature must have units of Kelvins. In our case the volume is fine, we must change the temperature from centigrade to Kelvins.

We do this by adding $273K$ $"273K"$ to the given temperatures. Thus the first temp. is

$273K + {8.0}^{\circ} C = 281K$ $"273K" + 8.0^@"C" = "281K"$

and the second temp. is

$273K + {60.0}^{\circ} C = 333K$ $"273K" + 60.0^@"C" = "333K"$

Now we just rearrange the equation and plug and chug.

$V_{2} = \frac{T_{2} \cdot V_{1}}{T_{1}}$ $V_2 = (T_2 * V_1)/(T_1)$

$V_2 = (333cancel("K") * "6.0 L") / (281 cancel("K"))$

$V_2 = "7.11 L"$

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Question #5bfc7

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