Question

Someone hands you a sample of carbon monoxide with a pressure of 4.11kPa, a volume of 5.22L, and a temperature of 405.4K If you double the pressure and reduce the temperature to 124.8K. What will the resulting volume of the gas be in liters?

Answer 1

We could use the combined law of gases:

`(P_1\cdot V_1)/(T_1)=(P_2\cdot V_2)/(T_2)`

Replacing the values of the problem, we got that:

`\begin{gathered} P_1=4.11kPa \\ V_1=5.22L \\ T_1=405.4K \\ P_2=8.22kPa \\ T_2=124.8K \\ V_2=?_{} \end{gathered}`

Now, we could solve the equation given for V2 with all these given values:

`\begin{gathered} (P_1\cdot V_1)/(T_1)=(P_2\cdot V_2)/(T_2)\to V_2=(P_1\cdot V_1\cdot T_2)/(T_1\cdot P_2)_{} \\ \\ V_2=(4.11kPa\cdot5.22L\cdot124.8K)/(405.4K\cdot8.22kPa) \\ \\ V_2=0.80L^{}_{} \end{gathered}`

Therefore, the new volume of the gas is 0.80L.