Question

30. A rigid plastic container holds 1.00 L methane gas at 0.9
atm pressure when the temperature is 22.0°C. How much
more pressure will the gas exert if the temperature is raised
to 44.6°C?​

Answer 1

Answer: The new pressure exerted by the gas is 0.646 atm

Explanation:

To calculate the final temperature of the system, we use the equation given by Gay-Lussac Law. This law states that pressure of the gas is directly proportional to the temperature of the gas at constant pressure.

Mathematically,

$\frac{P_1}{T_1}=\frac{P_2}{T_2}$

where,

$P_1\text{ and }T_1$ are the initial pressure and temperature of the gas.

$P_2\text{ and }T_2$ are the final pressure and temperature of the gas.

We are given:

$P_1=0.9atm\\T_1=22^oC=[22+273]K=295K\\P_2=?\\T_2=44.6^oC=[44.6+273]K=317.6K$

Putting values in above equation, we get:

$\frac{0.6atm}{295K}=\frac{P_2}{317.6}\\\\T_2=\frac{317.6\times 0.6}{295}=0.646atm$

Hence, the new pressure exerted by the gas is 0.646 atm

Answer 2

The more pressure that will the gas exert if the temperature is raised

to 44.6° C is 0.646 atm.

What is Boyle's Law?

It is a gas law that states the pressure decrease with the increase in the pressure.

By the formula

$\dfrac{P_1}{T_1 } =\dfrac{P_2}{T_2}$

$\dfrac{ 0.9}{295 } =\dfrac{P_2}{317.6} = 0.646 atm.$

Thus, the pressure that will the gas exert if the temperature is raised

to 44.6° C is 0.646 atm.

Learn more about Boyle's Law

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