Question

question 9 of 18
a sample of an ideal gas has a volume of 3.15 l at 11.80 °c and 1.50 atm. what is the volume of the gas at 24.40 °c and 0.992 atm?
v =

Explanation:

Step1: Convert temperatures to Kelvin

The formula to convert Celsius to Kelvin is \(T(K)=T(^{\circ}C)+273.15\).
For the initial temperature \(T_1 = 11.80^{\circ}C\), \(T_1=11.80 + 273.15=284.95\space K\).
For the final temperature \(T_2 = 24.40^{\circ}C\), \(T_2=24.40+ 273.15 = 297.55\space K\).

Step2: Use the combined gas law

The combined gas law is \(\frac{P_1V_1}{T_1}=\frac{P_2V_2}{T_2}\).
We are given \(P_1 = 1.50\space atm\), \(V_1=3.15\space L\), \(T_1 = 284.95\space K\), \(P_2=0.992\space atm\), \(T_2 = 297.55\space K\), and we need to find \(V_2\).
Rearrange the formula for \(V_2\): \(V_2=\frac{P_1V_1T_2}{P_2T_1}\).
Substitute the values: \(V_2=\frac{1.50\times3.15\times297.55}{0.992\times284.95}\).
First, calculate the numerator: \(1.50\times3.15\times297.55=(1.50\times3.15)\times297.55 = 4.725\times297.55=1406.92375\).
Then, calculate the denominator: \(0.992\times284.95 = 282.6604\).
Now, \(V_2=\frac{1406.92375}{282.6604}\approx5.0\space L\).

Answer:

\(5.0\space L\)