Question

If the temperature of the helium balloon were increased from 30°C to 35°C and the volume of the balloon only expanded from 0.47L to 1.55L, how much pressure would be inside the balloon?

Answer 1

Answer:

$\large \boxed{\text{0.31 times the original pressure}}$

Explanation:

To solve this problem, we can use the Combined Gas Laws:

$\dfrac{p_{1}V_{1} }{n_{1}T_{1}} = \dfrac{p_{2}V_{2} }{n_{2}T_{2}}$

Data:

p₁ = p₁; V₁ = 0.47 L; n₁ = n₁;  T₁ = 30 °C

p₂ = ?;  V₂ = 1.55 L; n₂ = n₁; T₂ = 35 °C

Calculations:

(a) Convert the temperatures to kelvins

T₁ = (30 + 273.15)K = 303.15 K

T₂ = (35 + 273.15)K = 308.15 K

(b) Calculate the new pressure

$\begin{array}{rcl}\dfrac{p_{1}V_{1}}{n_{1} T_{1}} & = & \dfrac{p_{2}V_{2}}{n_{2} T_{2}}\\\\\dfrac{p_{1}\times \text{0.47 L}}{n _{1}\times \text{303.15 K}} & = &\dfrac{p_{2}\times\text{1.55 L}}{n _{1}\times \text{308.15 K}}\\\\1.55 \times 10^{-3}p_{1} & = &5.030 \times 10^{-3}{p_{2}}\\p_{2} & = & \dfrac{1.55}{5.030} \times p_{1}\\\\ & = & 0.31p_{1}\\\end{array}\\\text{The new pressure is \large \boxed{\textbf{0.31 times the original pressure}} }$