Question

Se tienen 5 cm3 de aire encerrado en una jeringa, siendo p = 760

Answer 1

Answer:

a. 840 mmHg; b. 760 mmHg  

Explanation:

The only variables are the pressure and the volume, so we can use Boyle's Law.

p₁V₁ = p₂V₂

a. Pressure at one-tenth less volume

Data:

p₁ = 760 mmHg; V₁ = 5 cm³

p₂ = ?;                 V₂ = V₁ - 10 %

Calculations:

(i) Calculate V₂

0.1V₁ = 0.1 × 5 cm³        = 0.5 cm³

   V₂ = 5 cm³ - 0.5 cm³ = 4.5 cm³

(ii) Calculate p₂

$\begin{array}{rcl}p_{1}V_{1} & = & p_{2}V_{2}\\\text{760 mmHg} \times \text{5 cm}^{3} & = & p_{2} \times \text{4.5 cm}^{3}\\\text{3800 mmHg} & = & 4.5p_{2}\\p_{2} & = & \dfrac{\text{3800 mmHg}}{4.5}\\\\& = &\textbf{840 mmHg}\\\end{array}$

$\text{The new pressure of the gas is \textbf{840 mmHg} }$

b. Pressure at one-half the volume

Data:

p₁ = 760 mmHg; V₁ = 5 cm³

p₂ = ?;                  V₂ = ½V₁  

Calculations:

(i) Calculate V₂

V₂ = ½V₁ = ½ × 5 cm³ = 2.5 cm³

(ii) Calculate p₂

$\begin{array}{rcl}p_{1}V_{1} & = & p_{2}V_{2}\\\text{760 mmHg} \times \text{5 cm}^{3} & = & p_{2} \times \text{2.5 cm}^{3}\\\text{3800 mmHg} & = & 2.5p_{2}\\p_{2} & = & \dfrac{\text{3800 mmHg}}{2.5}\\\\& = &\text{1520 mmHg}\\\end{array}$

(iii) Calculate the increase in pressure

Δp = p₂ - p₁ = 1520 mmHg - 760 mmHg = 760 mmHg

The pressure must increase by 760 mmHg.