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**Hint:**Use the ideal gas equation $PV = nRT$ to find the initial volume. Apply the first law of thermodynamics of heat transfer at constant pressure $Q = n{C_P}({T_2} - {T_1})$ and calculate the final temperature. Now, substitute the known data in Charles’ law expression $\dfrac{{{V_1}}}{{{V_2}}} = \dfrac{{{T_1}}}{{{T_2}}}$ to find the final volume.

**Complete step-by-step solution**

The equation which relates the pressure , volume and temperature of the given state of an ideal gas is known as the ideal gas equation or equation of state.

$PV = nRT$

Given that

\[

{C_v} = 2.5R \\

{T_1} = {\text{ }}{0^o}C{\text{ }} = 273.15K \\

P = 1{\text{ }}atm \\

\]

Substitute in the ideal gas equation.

$

PV = nR{T_1} \\

1 \times V = 1 \times 0.082 \times 273.15 \\

V = 22.4L \\

$

We know that,

$

{C_P} = R + {C_V} \\

{C_P} = R + 2.5R \\

{C_P} = 3.5R \\

$

The heat transferred Q at constant pressure is 13200J.

From first law of thermodynamics,

$

Q = n{C_P}({T_2} - {T_1}) \\

13200 = 1 \times 3.5 \times 8.314 \times ({T_2} - 273.15) \\

{T_2} = 726.77K \\

$

\[{V_1}, {\text{ }}{V_2}\] and \[{T_1},{\text{ }}{T_2}\] are the initial and final volume and temperature respectively.

According to Charles law, for the pressure remaining constant, the volume of the given mass of a gas is directly proportional to its absolute temperature.

$

V \propto T \\

\dfrac{{{V_1}}}{{{V_2}}} = \dfrac{{{T_1}}}{{{T_2}}} \\

{V_2} = \dfrac{{726.77 \times 22.4}}{{273.15}} \\

{V_2} = 60L \\

$

**Hence, the final volume is 60 L and the correct option is C.**

**Note:**Boyle’s law states that for a given mass of an ideal gas at constant temperature , the volume is inversely proportional to the pressure.

${P_1}{V_1} = {P_2}{V_2}$

Gay-Lussac’s law states that the volume remaining constant, the pressure of a given mass of a gas is directly proportional to its absolute temperature.

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