QuestionA sealed can of gas has a pressure of 2.0 atm at 300 K . If the can is heated to 600 K , what will the pressure become (assuming constant volume)?

Studdy Solution

STEP 1

1. The initial pressure of the gas is 2.0 atm.
2. The initial temperature of the gas is 300 K.
3. The final temperature of the gas is 600 K.
4. The volume of the gas is constant.
5. The gas behaves ideally, and we can use the ideal gas law.

STEP 2

1. Identify the relevant gas law.
2. Set up the equation using the gas law.
3. Solve for the final pressure.

STEP 3

Identify the relevant gas law. Since the volume is constant, we use the pressure-temperature relationship of the ideal gas law, which is given by Gay-Lussac's Law:
$\frac{P_1}{T_1} = \frac{P_2}{T_2}$
where $P_1$ and $T_1$ are the initial pressure and temperature, and $P_2$ and $T_2$ are the final pressure and temperature.

STEP 4

Set up the equation using the gas law. Substitute the known values into the equation:
$\frac{2.0 \, \text{atm}}{300 \, \text{K}} = \frac{P_2}{600 \, \text{K}}$

STEP 5

Solve for the final pressure $P_2$ . Cross-multiply to solve for $P_2$ :
$P_2 = \frac{2.0 \, \text{atm} \times 600 \, \text{K}}{300 \, \text{K}}$
Simplify the expression:
$P_2 = \frac{1200 \, \text{atm} \cdot \text{K}}{300 \, \text{K}}$
$P_2 = 4.0 \, \text{atm}$
The final pressure of the gas is:
$\boxed{4.0 \, \text{atm}}$

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