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

New volume =
`1.69* 10^6` L

Step-by-step explanation:

Using Charle's law

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

Given ,

V₁ =
`1.27* 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:

`(1.27* 10^6)/(284.15)=(V_2)/(378.15)`

`V_2=(1.27* 10^6\cdot \:378.15)/(284.15)`

New volume =
`1.69* 10^6` L