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

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

Step-by-step explanation:

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

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