Question

A flexible container is put in a deep freeze. its original volume is 3.00 m3 at 25.0°c. after the container cools, it has shrunk to 2.00 m3. its new temperature in degrees celsius is

Answer 1

To solve this we assume that the gas inside the balloon is an ideal gas. Then, we can use the ideal gas equation which is expressed as PV = nRT. At a constant pressure and number of moles of the gas the ratio T/V is equal to some constant. At another set of condition of temperature, the constant is still the same. Calculations are as follows:

T1 / V1 = T2 / V2

T2 = T1 x V2 / V1 

T2 = 25 x 3 / 2

T2 = 37.5 degrees Celsius

Answer 2

Answer : The new temperature will be, $-74.4^oC$

Explanation :

Charles' Law : This law states that volume of gas 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)

or,

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

where,

$V_1$ = initial volume of gas = $3.00m^3$

$V_2$ = final volume of gas = $2.00m^3$

$T_1$ = initial temperature of gas = $25^oC=273+25=298K$

$T_2$ = final temperature of gas = ?

Now put all the given values in the above formula, we get the final temperature of gas.

$\frac{3.00m^3}{298K}=\frac{2.00m^3}{T_2}$

$T_2=198.6K=198.6-273=-74.4^oC$

conversion used : $^oC=K-273$

Therefore, the temperature will be, $-74.4^oC$