Question

Calculate the Kelvin temperature to which 10.0 L of a gas at 27 °C would have to be heated to change the volume to 12.0 L. Units of Kelvin are assumed here for the answer you enter so just put the number value.

Answer 1

360.18 K is the Kelvin temperature to which 10.0 L of a gas at 27 °C would have to be heated to change the volume to 12.0 L.

Explanation:

Data given:

Initial volume of the gas, V1 = 10 litres

Initial temperature of the gas, T1 = 27° C or 273.15+27 = 300.15 K

Final volume of the gas obtained, V2 = 12 Litres

final temperature to obtain the above volume, T2 =?

temperature value in Kelvin

Applying Charles' Law to the data given,

$\frac{V1}{T1} =\frac{V2}{T2}$

rearranging the equation to get T2,

T2 = $\frac{V2T1}{V1}$

Putting the values in the equation:

T2 =$\frac{ 12 X 300.15}{10}$

T2 = 360.18 K

The gas will be heated at the temperature of 360.18 K to get its volume changed to 12 litres.

Answer 2

The temperature will be "360 K".

According to the question,

Volume,

• $V_1 = 10.0 \ L$
• $V_2 = 12.0 \ L$

Temperature,

• $T_1 = 300 \ K$
• $T_2 = \ ?$

By using the relation, we get

→ $PV= nRT$

then,

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

→   $T_2= \frac{T_1\times V_2}{V_1}$

By substituting the values,

         $= \frac{300\times 12.0}{10}$

         $= 360 \ K$

Thus the above answer is correct.

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