Question

a sealed container holds a gas at a pressure of 1.5 atm and a temperature of 300 k. if the temperature is increased to 450 k while the volume remains constant, what will be the new pressure of the gas? the new pressure would be select one select one

Explanation:

Step1: Identify the gas law

Since volume is constant, we use Gay - Lussac's Law, which is $\frac{P_1}{T_1}=\frac{P_2}{T_2}$, where $P_1$ is the initial pressure, $T_1$ is the initial temperature, $P_2$ is the final pressure, and $T_2$ is the final temperature.

Step2: List the given values

We know that $P_1 = 1.5\ atm$, $T_1=300\ K$, and $T_2 = 450\ K$. We need to find $P_2$.

Step3: Rearrange the formula to solve for $P_2$

From $\frac{P_1}{T_1}=\frac{P_2}{T_2}$, we can rearrange it to $P_2=\frac{P_1\times T_2}{T_1}$.

Step4: Substitute the values into the formula

Substitute $P_1 = 1.5\ atm$, $T_1 = 300\ K$, and $T_2=450\ K$ into the formula: $P_2=\frac{1.5\ atm\times450\ K}{300\ K}$.
First, calculate the numerator: $1.5\times450 = 675$. Then divide by the denominator: $\frac{675}{300}=2.25$. So $P_2 = 2.25\ atm$.

Answer:

The new pressure of the gas is $2.25\ atm$.