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

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,

`(V_1)/(T_1)=(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.

`(3.00m^3)/(298K)=(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`

Answer 2

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