The volume of a gas is 10.2 L when the temperature is 9.5 $^{o}$ $"^o$ C. If the volume is increased to 64.9 L without changing the pressure what is the new temperature?

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Answer 1 · 2016-11-20T13:26:02

The new temperature is $= 1524.5$ $=1524.5$ ºC

we need to use Charles' law

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

The temperatures are in Kelvin (K)

$K=ºC+273$

$V_1=10.2l$

$T_1=9.5ºC=9.5+273=282.5K$

$V_2=64.9l$

so, $T_2=V_2*T_1/V_1=64.9*282.5/10.2=1797.5K$

$T_2=1797.5-273=1524.5ºC$

The volume of a gas is 10.2 L when the temperature is 9.5"^oo"^oC. If the volume is increased to 64.9 L without changing the pressure what is the new temperature?