Question

the volume of a sample of oxygen is 200.0 ml when the pressure is 3.000 atm, and the temperature is 37.0°c. what is the new temperature if the volume increases to 400.0 ml and the pressure decreases to 2.000 atm?
a 413.0°c
b 209.8°c
c 49.33°c
d 140.3°c

Explanation:

Step1: Convert initial temperature to Kelvin

$T_1 = 37.0 + 273.15=310.15\ K$

Step2: Apply the combined - gas law formula $\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}$

We know $P_1 = 3.000\ atm$, $V_1 = 200.0\ mL$, $T_1 = 310.15\ K$, $P_2 = 2.000\ atm$, $V_2 = 400.0\ mL$. Rearranging for $T_2$ gives $T_2=\frac{P_2V_2T_1}{P_1V_1}$.

Step3: Substitute values into the formula

$T_2=\frac{2.000\ atm\times400.0\ mL\times310.15\ K}{3.000\ atm\times200.0\ mL}=413.53\ K$

Step4: Convert final temperature back to Celsius

$T_2 = 413.53 - 273.15 = 140.38\approx140.3^{\circ}C$

Answer:

D. $140.3^{\circ}C$