1.0 mole of a gas is enclosed in a 12.3 l cylinder with a moveable piston at 300. k and 2.0 atm. half of the…

Explanation:

Step1: Recall the ideal - gas law

The ideal - gas law is $PV = nRT$, where $P$ is pressure, $V$ is volume, $n$ is the number of moles, $R$ is the ideal - gas constant ($R=0.0821\ L\cdot atm/(mol\cdot K)$), and $T$ is temperature in Kelvin. Since the pressure is constant ($P_1 = P_2$), we can use the combined gas - law in the form $\frac{V_1}{n_1T_1}=\frac{V_2}{n_2T_2}$.

Step2: Identify the initial and final values

Initial values: $n_1 = 1.0\ mol$, $V_1 = 12.3\ L$, $T_1 = 300\ K$. Final values: $n_2 = 0.50\ mol$, $T_2 = 900\ K$, and we need to find $V_2$.

Step3: Rearrange the combined - gas law formula to solve for $V_2$

From $\frac{V_1}{n_1T_1}=\frac{V_2}{n_2T_2}$, we can cross - multiply to get $V_2=\frac{V_1n_2T_2}{n_1T_1}$.

Step4: Substitute the values into the formula

$V_2=\frac{12.3\ L\times0.50\ mol\times900\ K}{1.0\ mol\times300\ K}$. First, calculate the numerator: $12.3\times0.50\times900 = 12.3\times450=5535$. Then, calculate the denominator: $1.0\times300 = 300$. $V_2=\frac{5535}{300}=18.45\ L$.

Answer:

$18.45\ L$