Question

A gas container has a volume of 446.9 with a temp of 14c. When the volume is decreased to 238.7l the new temp is what

Answer 1

Answer:

$\frac{V _{1}}{T _{1}} = \frac{V _{2}}{T _{2} } \\ \frac{446.9}{(14 + 273)} = \frac{238.7}{T _{2} } \\ {T _{2}} = \frac{238.7 \times 287}{446.9} \\ {T _{2}} = 153.3 \: kelvin \\ = 119.7 \degree \: c$