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.

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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.

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Answer: The final pressure is 12.6 atmExplanation:The combined gas equation is, [tex]\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}[/tex] where, [tex]P_1[/tex] = initial pressure of gas = 5.6 atm [tex]P_2[/tex] = final pressure of gas = ? [tex]V_1[/tex] = initial volume of gas = v [tex]V_2[/tex] = final volume of gas = [tex]v-\frac{50}{100}v=0.5v[/tex] [tex]T_1[/tex] = initial temperature of gas = [tex]-19.0^0C=(-19+273)K=254K[/tex] [tex]T_2[/tex] = final temperature of gas = [tex]12.0^0C=(12.0+273)K=285K[/tex] Now put all the given values in the above equation, we get: [tex]\frac{5.6\times v}{254}=\frac{P_2\times 0.5v}{285}[/tex] [tex]P_2=12.6atm[/tex] The final pressure is 12.6 atm

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