A container with a volume of $42 L$ $42 L$ contains a gas with a temperature of $120^{o} K$ $120^o K$ . If the temperature of the gas changes to $75^{o} K$ $75 ^o K$ without any change in pressure, what must the container's new volume be?

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Answer 1 · 2016-04-01T21:39:00

$V_{2} = 26, 25 L$ $V_2=26,25 L$

$given data:$ $"given data:"$
$V_{1} = 42 L$ $V_1=42 L$
$T_{1} = 120^{o} K$ $T_1=120^o K$
$T_{2} = 75^{o} K$ $T_2=75^o K$
$V_{2} = ?$ $V_2=?$

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

$\frac{42}{120} = \frac{V_{2}}{75}$ $42/120=V_2/75$

$V_{2} = \frac{42 \cdot 75}{120}$ $V_2=(42*75)/120$

$V_{2} = 26, 25 L$ $V_2=26,25 L$

A container with a volume of 42 L42L42 L contains a gas with a temperature of 120^o K120oK120^o K. If the temperature of the gas changes to 75 ^o K75oK75 ^o K without any change in pressure, what must the container's new volume be?