Question

A hot air balloon contains 5.30 kL of helium gas when the temperature is 12°C. At what temperature will the balloon's volume have increased to 6.00 kL? (Remember to convert to Kelvin and then back to Celcius.)

(A) 50 °C
(B) -21 °C

Answer 1

(A) 50 °C

Step-by-step explanation:

Ideal gas law:

PV = nRT

If pressure is held constant, then:

V₁ / T₁ = V₂ / T₂

Substituting values:

(5.30 kL) / (12 + 273 K) = (6.00 kL) / T

T = 323 K

T = 50 °C

Answer 2

Option (A) 50 °C

Step-by-step explanation:

Thinking process:

Let the volume of the hot air be
`V_(1)` 5.30 k l = 5 300 l = 5.3 m³

The initial temperature will be 12 ° (12 + 273.15) = 285.15 K

The later volume,
`V_(2)` = 6 Kl = 6 000 l = 6 m³

The equation will be:

`(P_(1)V_(1) )/(T_(1) ) = (P_(2)V_(2) )/(T_(2) )`

If the pressure is the same, then:

`(5.3)/(285.15) = (6)/(T_(2) )`

Solving for
`T_(2)` gives:

T₂ = 322.8

= 49.66

= 50°