Question

For many purposes we can treat methane as an ideal gas at temperatures above its boiling point of . Suppose the temperature of a sample of methane gas is lowered from ⁢ to , and at the same time the pressure is changed. If the initial pressure was and the volume increased by , what is the final pressure? Round your answer to significant digits.

Answer 1

Answer:

1.3 × 10 atm

Explanation:

There is some info missing. I think this is the original question.

For many purposes we can treat methane (CH₄) 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 -19.0 °C to 12.0 °C, and at the same time the pressure is changed. If the initial pressure was 5.6 atm and the volume decreased by 50.0%, what is the final pressure? Round your answer to the correct number of significant digits. atm x10 ?

Given data

T₁ = -19.0°C + 273.15 = 254.2 K

T₂ = 12.0°C + 273.15 = 285.2 K

V₁ (unknown)

V₂ = 50.0% V₁ = 0.500 V₁

P₁ = 5.6 atm

P₂ (what we are looking)

We can find the the final pressure P₂ using the combined law gas.

$\frac{P_{1}.V_{1}}{T_{1}} =\frac{P_{2}.V_{2}}{T_{2}} \\\frac{P_{1}.V_{1}}{T_{1}} =\frac{P_{2}.0.500V_{1}}{T_{2}} \\\frac{P_{1}.}{T_{1}} =\frac{P_{2}.0.500}{T_{2}}\\P_{2}=\frac{P_{1}.T_{2}}{0.500T_{1}} =\frac{5.6atm \times 285.2K}{0.500 \times254.2} =13 atm$