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

Answer:

(A) 50 °C

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

Answer:

Option (A) 50 °C

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:

$\frac{P_{1}V_{1} }{T_{1} } = \frac{P_{2}V_{2} }{T_{2} }$

If the pressure is the same, then:

$\frac{5.3}{285.15} = \frac{6}{T_{2} }$

Solving for $T_{2}$ gives:

T₂ = 322.8

   = 49.66

   = 50°