A container with a volume of $12 L$ $12 L$ contains a gas with a temperature of $240^{o} C$ $240^o C$ . If the temperature of the gas changes to $420^{o} K$ $420 ^o K$ without any change in pressure, what must the container's new volume be?

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Answer 1 · 2018-01-06T06:55:06

The new volume is $= 9.82 L$ $=9.82L$

Apply Charles'Law

$\frac{V_{1}}{T_{1}} = \frac{V_{2}}{T_{2}}$ $V_1/T_1=V_2/T_2$ $at constant pressure$ $"at constant pressure"$

The initial volume is $V_{1} = 12 L$ $V_1=12L$

The initial temperature is $T_{1} = 240 + 273 = 513 K$ $T_1=240+273=513K$

The final temperature is $T_{2} = 420 K$ $T_2=420K$

The final volume is

$V_{2} = \frac{T_{2}}{T_{1}} \cdot V_{1} = \frac{420}{513} \cdot 12 = 9.82 L$ $V_2=T_2/T_1*V_1=420/513*12=9.82L$

A container with a volume of 12 L12L12 L contains a gas with a temperature of 240^o C240oC240^o C. If the temperature of the gas changes to 420 ^o K420oK420 ^o K without any change in pressure, what must the container's new volume be?