Aquantity of gas has a volume of 0.20 cubic meter and an absolute temperature of 333 degrees kelvin. when the temperature of the gas is raised to 533 degrees kelvin, what is the new volume of the gas? (assume that there's no change in pressure)

V1 = 0.2m3
v2=?
t1=333
t2 = 533
charles law: v1/t1=v2/t2

v2=v1/t1 x t2
    =0.2/333 x533

v2 = 0.32 cubic meter

Initial velocity,  u = 4.9 m/sfinal velocity = v m/sinitial height,  h₁ =  245 m    final height, h₂ = 0a = -9.8  m/s²time taken = t secondss = ut + 0.5at²⇒  (h₂-h₁) = ut + 0.5at²⇒  0-245 = 4.9t + 0.5×(-9.8)×t²⇒ -245 = 4.9t - 4.9t²⇒ 4.9t² -4.9t -245 =0solving it, we get   t = 7.59sv  = u +  at = 4.9 -9.8×7.59 = 4.9 - 74.38 = -69.48 m/svelocity is  69.48 m/s downward

Answeroutward from: source;