Question

A closed container contains 0.40 moles of argon gas at 25 °C. If the temperature is doubled while keeping the volume constant, what is the new pressure of the argon gas?

A. No change in pressure
B. 0.40 atm
C. 0.80 atm
D. 1.20 atm

Answer 1

The correct answer is A. No change in pressure.

Step-by-step explanation:

According to the ideal gas law, \(PV = nRT\), where \(P\) is pressure, \(V\) is volume, \(n\) is the number of moles, \(R\) is the gas constant, and \(T\) is the temperature in Kelvin. When the volume is constant (as mentioned in the question), the relationship between pressure and temperature becomes \(P_1/T_1 = P_2/T_2\).

In this scenario, the initial temperature is 25 °C, which is equivalent to 298 K. If the temperature is doubled, the new temperature (\(T_2\)) becomes 2 times 298 K, which is 596 K. Since the volume is constant, the initial and final pressures (\(P_1\) and \(P_2\)) are directly proportional to their respective temperatures.

\[ \frac{P_1}{T_1} = \frac{P_2}{T_2} \]

\[ \frac{P_1}{298} = \frac{P_2}{596} \]

Solving for \(P_2\):

\[ P_2 = \frac{P_1 \times 596}{298} \]

\[ P_2 = 2 \times P_1 \]

This indicates that the pressure doubles when the temperature is doubled while keeping the volume constant. Therefore, the new pressure is 2 times the initial pressure. As none of the given options reflect this change, the correct answer is A. No change in pressure. This explanation provides a step-by-step approach to understanding the relationship between pressure and temperature in an ideal gas when the volume is held constant.

Answer 2

A closed container contains 0.40 moles of argon gas at 25 °C. If the temperature is doubled while keeping the volume constant what is the new pressure of the argon gas?

C. 0.80 atm

Step-by-step explanation:

When the temperature of a gas is doubled while keeping the volume constant according to Gay-Lussac's Law the pressure of the gas also doubles. The relationship between pressure P and temperature T at constant volume is given by the equation
`\(P_1/T_1 = P_2/T_2\)` where
`\(P_1\` and
`(T_1\`) are the initial pressure and temperature and
`\(P_2\) and \(T_2\)` are the final pressure and temperature respectively.

In this case, the initial temperature
`\(T_1\) is 25 °C` which is equivalent to 298 K. Doubling the temperature we get
`\(T_2 = 2 * 298 = 596\) K`. Since the volume is constant we can set up the proportion:

`\[ (P_1)/(T_1) = (P_2)/(T_2) \]`

Substituting the known values:

`\[ (P_1)/(298) = (P_2)/(596) \]`

Solving for
`\(P_2\)`:

`P_2 = P_1 * (596)/(298) \]\[`

Given that P_1 is not specified, it cancels out in the ratio, leaving
`P_2 = 2 * P_1\`. Therefore, the final pressure
`(\(P_2\))` is double the initial pressure
`(\(P_1\))` resulting in a new pressure of 0.80 atm.

In conclusion when the temperature of a gas in a closed container is doubled while the volume is held constant the pressure of the gas also doubles. Applying the appropriate formula and substituting the known values allow us to determine the final pressure of the argon gas resulting in the correct answer of 0.80 atm.