Question

at 10. °c, 20. g of oxygen gas exerts a pressure of 2.1 atm in a rigid, 7.0 l cylinder. assuming ideal behavior, if the temperature of the gas was raised to 40. °c, which statement indicates the new pressure and explains why? a 1.9 atm, because the pressure p decreases by the proportion 283/313. b 2.3 atm, because the pressure p increases by the proportion 313/283. c 0.52 atm, because the pressure p decreases by the proportion 10/40. d 8.4 atm, because the pressure p increases by the proportion 40/10.

Explanation:

Step1: Convert temperatures to Kelvin

Initial temperature $T_1=(20 + 273)\text{ K}=293\text{ K}$, final temperature $T_2=(40+ 273)\text{ K}=313\text{ K}$. Initial pressure $P_1 = 2.1\text{ atm}$.

Step2: Apply Gay - Lussac's Law

Gay - Lussac's Law is $\frac{P_1}{T_1}=\frac{P_2}{T_2}$, so $P_2=\frac{T_2}{T_1}\times P_1$.

Step3: Calculate the final pressure

$P_2=\frac{313}{293}\times2.1\text{ atm}\approx2.3\text{ atm}$. Since $T_2>T_1$, pressure increases. The ratio of pressures is $\frac{P_2}{P_1}=\frac{313}{293}$, which means pressure increases by the proportion $\frac{313}{293}$.

Answer:

B. 2.3 atm, because the pressure $P$ increases by the proportion $\frac{313}{293}$