Question

A hot air balloon filled with 1.27x10^6L of an ideal gas on a cool morning of 11 degrees C. the air is heated to 105 degrees C. What is the volume of the air in the ballon after it is heated?Assume that none of the gas escapes from the balloon.L=_______

Answer 1

Answer:

New volume = $1.69\times 10^6$ L

Explanation:

Using Charle's law  

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

Given ,  

V₁ = $1.27\times 10^6$ L

V₂ = ?

T₁ = 11 °C

T₂ = 105 °C  

The conversion of T( °C) to T(K) is shown below:

T(K) = T( °C) + 273.15  

So,  

T₁ = (11 + 273.15) K = 284.15 K  

T₂ = (105 + 273.15) K = 378.15 K  

Using above equation as:

$\frac{1.27\times 10^6}{284.15}=\frac{V_2}{378.15}$

$V_2=\frac{1.27\times 10^6\cdot \:378.15}{284.15}$

New volume = $1.69\times 10^6$ L