Question

At 10.°C, 20.g of oxygen gas exerts a pressure of 2.1atm in a rigid, 7.0L cylinder. Assuming ideal behavior, if the temperature of the gas was raised to 40.°C, which statement indicates the new pressure and explains why?

Answer 1

The new pressure of the gas if the ideal gas situation is assumed is; P2 = 2.323atm.

The statement which explains why is the Gay lussac's law; otherwise known as the Amonton's law.

qqThe general ideal gas equation is;

• PV = nRT

The ideal gas las is however modified in the Amonton's law which states that;

• At constant volume, the pressure of a given mass of a gas is directly proportional to its temperature.

Therefore;

• P/T = constant.

Ultimately, P1/P2 = T1/T2

where:

• P1 = 2.1atm

• P2 = ?

• T1 = 10⁰c = 283 K

• T2 = 40°c = 313 K

P2 = (2.1 × 313)/283

P2 = 2.323 atm.

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Answer 2

Ideal gas is a hypothetical gas in which all the collision between the particles are perfectly elastic and have no intermolecular attractive forces. The ideal gas new pressure in the given situation is P$_2$ = 2.323 atm.

Ideal gas equation is used for the approximation behaviour of the general gases. The ideal gas is defined as:

1. the gas in which particles or molecules do not have attractive and repulsive forces.
2. the gas have no volume or space.

Given that,

• T$_1$ or temperature of oxygen = 10-degrees celcius
• Pressure or P$_1$ exerted by oxygen = 2.1 atm
• T$_2$ or temperature raised = 40-degrees celcius
• P$_2$ or pressure after raise in temperature = ?

Now, from the ideal gas equation, :

PV=nRT

At constant volume, the pressure of given mass of gas is directly proportional to the temperature, such that:

• $\dfrac{\text P}{\text T}&=\text {constant}$

• Likewise, $\dfrac{\text P_1}{\text P_2}&=\dfrac{\text T_1}{\text T_2}$

where,

• $\text P_1 &= 2.1 \text {atm}\\\text P_2 &= ?\\\text T_1 &= 283 \;\text K &= 10^0 \text C\\\text T_2 &= 313\;\text K &=40^0 \text C$

Now, substituting the values in the above equation:

• $\begin{aligned}\dfrac{\text 2.1}{\text P_2}&=\dfrac{\text 283}{\text 313}\\\text P_2&= \dfrac{2.1\times 313}{283}\\\text P_2&=2.323\text{atm}\end{aligned}$

Therefore, the pressure after raising the temperature of the gas will be 2.323 atm.

To know more about ideal gas equation, refer to the following link:

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