Question

A hot air balloon is filled with 1.13 × 10 6 L 1.13×106 L of an ideal gas on a cool morning ( 11 ∘ C ) . 11 ∘C). The air is heated to 127 ∘ C . 127 ∘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.L=_________.

Answer 1

New volume =
`1.59* 10^6` L

Step-by-step explanation:

Using Charle's law

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

Given ,

V₁ =
`1.13* 10^6` L

V₂ = ?

T₁ = 11 °C

T₂ = 127 °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 = (127 + 273.15) K = 400.15 K

Using above equation as:

`(1.13* 10^6)/(284.15)=(V_2)/(400.15)`

`V_2=(1.13* 10^6\cdot \:400.15)/(284.15)`

New volume =
`1.59* 10^6` L