Question #e05ec

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Question #e05ec

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Answer 1 · 2016-10-25T17:45:48

The new volume is 27.4 liters

Explanation:

According to Charles Law's volume and temperature is a direct relationship. As the temperature decreases the volume also decreases.

The key is to solve the equation in terms of degrees K or absolute temperature.

The equation is

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

$V_{1} = 72.3 l$ $V_1 = 72.3 l$
$T_{1} = 307 k^{o}$ $T_1 = 307 k^o$ ( $36 C^{o} + 273 C^{o}$ $36C^o + 273 C^o$ ) $0^{o} C = 273^{o} K$ $0^oC = 273^oK$
$T_{2} = 117 k^{o}$ $T_2 = 117 k^o$ ( $- 156^{o} C - 273 C^{o})$ $-156^oC - 273C^o)$

$\frac{72.3}{307} = \frac{V_{2}}{117}$ $72.3/307 = V_2/117$ solving this gives

$V_{2} = 27.4 l$ $V_2 = 27.4 l$

Answer 2 · 2016-10-25T18:04:54

$27.4 l$ $27.4l$ , rounded to one decimal place.

Explanation:

As nothing has been mentioned about the pressure in the problem, it is treated to be constant.

According to Charles Law (Law of Volumes): When the pressure of a sample of gas is held constant, the Kelvin temperature and its volume are directly related.

Mathematically stated
$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$ $V_1/T_1 = V_2/T_2$ , at constant pressure. ........(1)

Given values are
$V_{1} = 72.3 l$ $V_1 = 72.3 l$
$T_{1} = 36 + 273 = 309 K$ $T_1 = 36+273=309 K$
$T_{2} = - 156 + 273 = 117 K$ $T_2 = -156 + 273=117K$
Inserting in the equation (1) we get

$\frac{72.3}{309} = \frac{V_{2}}{117}$ $72.3/309 = V_2/117$
$\Rightarrow V_{2} = \frac{72.3}{309} \times 117$ $=> V_2 = 72.3/309xx117$
$\Rightarrow V_{2} = 27.4 l$ $=> V_2 = 27.4l$ , rounded to one decimal place.

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Question #e05ec

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