250 ml of argon gas is held in a flexible vessel shown above. if the pressure changes to 12.0 atm, what is…

Answer:

A. 185 mL

Explanation:

Step1: Identify the combined - gas law formula

$ \frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

Step2: Convert temperatures to Kelvin

$T_1 = 25 + 273=298\ K$, $T_2=20 + 273 = 293\ K$

Step3: Substitute the given values

$P_1 = 10.0\ atm$, $V_1 = 250\ mL$, $P_2 = 12.0\ atm$, $T_1 = 298\ K$, $T_2 = 293\ K$ into $\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$. We get $\frac{10.0\ atm\times250\ mL}{298\ K}=\frac{12.0\ atm\times V_2}{293\ K}$

Step4: Solve for $V_2$

First, cross - multiply: $10.0\ atm\times250\ mL\times293\ K=12.0\ atm\times V_2\times298\ K$. Then $V_2=\frac{10.0\times250\times293}{12.0\times298}\ mL\approx185\ mL$