Question

For many purposes we can treat methane (CH) as an ideal gas at temperatures above its boiling point of - 161. °C. Suppose the temperature of a sample of methane gas is raised from 100.0 °C to 119.0 °C, and at the same time the pressure is decreased by 15.0%. increase Does the volume of the sample increase, decrease, or stay the same? decrease x 6 ? stays the same If you said the volume increases or decreases, calculate the percentage change in the volume. Round your answer to the nearest percent. 1%

Answer 1

Answer:

Explanation:

T₁ = 100 + 273 = 373K

T₂ = 273 + 119 = 392 K

V₁ = initial volume

V₂ = Final volume

P₁ = P₁

P₂ = .85P₁

Using gas law equation

$\frac{P_1\times V_1}{T_1} =\frac{P_2\times V_2}{T_2}$

$= \frac{P_1\times V_1}{373} =\frac{.85P_1\times V_2}{392}$

V₂ = 1.236 V₁

% increase in volume

= V₂-V₁ / V₁  x 100

= (1.236 V₁ - V₁ / V₁)x 100

= .236 x 100

= 23.6 % .