Question

At what temperature would the volume of a gas be 0.550 L if it had a volume of 0.432 L at –20.0 o C?

Answer 1

The temperature that  would  the volume of a gas  be 0.550l  if  it  had a volume of 0.432 L  at  -20.0  c is calculated  using the Charles law formula

that is   v1/T1=V2/T2
V1=0.550 l
t1=?
T2= -20 c +273 = 253 K
v2= 0.432 l

by  making T1  the subject of the formula  T1= V1T2/V2


T1=  (0.55lL x253)/  0.432 l = 322.11 K  or  322.11-273 = 49.11 C

Answer 2

Answer: 321.6 K

Explanation: Charles' Law: This law states that volume is directly proportional to the temperature of the gas at constant pressure and number of moles.

$V\propto T$    (At constant pressure and number of moles)

$\frac{V_1}{T_1}=\frac{V_2}{T_2}$

where,

$V_1$ = initial volume of gas = 0.432 L

$V_2$ = final volume of gas = 0.550 L

$T_1$ = initial temperature of gas = $-20^oC=273-20=253K$

$T_2$ = final temperature of gas = ?

Now put all the given values in the above equation, we get the final pressure of gas.

$\frac{0.432L}{253K}=\frac{0.550L}{T_2}$

$T_2=321.6K$

Therefore, the final temperature will be 321.6 K.