20.0 L of a gas originally at 22 °C and 1.05 atm pressure are compressed to 10.0 L.The new temperature is 57 °C.What is the new pressure?

Question

Quizplus · Accepted Answer

Using the combined gas law, $P_1V_1/T_1 = P_2V_2/T_2$, where $P$ is pressure, $V$ is volume, and $T$ is temperature in Kelvin. Convert temperatures to Kelvin: $T_1 = 22°C + 273 = 295K$ and $T_2 = 57°C + 273 = 330K$. Solve for $P_2$: $P_2 = P_1V_1T_2 / (V_2T_1) = (1.05 atm)(20.0 L)(330K) / (10.0 L)(295K) ≈ 2.35 atm$.

200 L of a Gas Originally at 22 °C and 1.05