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A hot air balloon is filled with 1.27 × 10 6 L of an ideal gas on a cool morning ( 11 ∘ C ) . The air is heated to 103 ∘ C . What is the volume of the air in the balloon after it is heated? Assume that none of the gas escapes from the balloon.

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Answer: Volume of the air in the balloon after it is heated is
1.68* 10^6L

Step-by-step explanation:

To calculate the final volume of the system, we use the equation given by Charles' Law. This law states that volume of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,


(V_1)/(T_1)=(V_2)/(T_2)

where,


V_1\text{ and }T_1 are the initial volume and temperature of the gas.


V_2\text{ and }T_2 are the final volume and temperature of the gas.

We are given:


V_1=1.27* 10^6\\T_1=11^oC=(11+273)K=284K\\V_2=?\\T_2=103^oC=(103+273)K=376K

Putting values in above equation, we get:


(1.27* 10^6)/(284K)=(V_2)/(376)\\\\V_2=1.68* 10^6L

Thus volume of the air in the balloon after it is heated is
1.68* 10^6L

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