Question

What is the final volume of a sample of neon under 5.6 atm of pressure at 250 k if it was initially at 6.7 atm or pressure, 500k, and was 18L?

Answer 1

Answer:

$V_2=10.8L$

Explanation:

Hello,

In this case, we use the combined idea gas equation that help us to understand the volume-pressure-temperature relationship as:

$\frac{P_1V_1}{T_1}= \frac{P_2V_2}{T_2}$

So, we solve for the final volume as shown below, considering the given data at the beginning of the experiment:

$V_2=\frac{P_1V_1T_2}{T_1P_2}=\frac{6.7atm*18L*250K}{500K*5.6atm}\\ \\V_2=10.8L$

Best regards.

Answer 2

Answer:

10.77 L

Explanation:

Initial pressure P1= 6.7atm

Initial temperature T1= 500K

Initial volume V1= 18L

Final pressure P2= 5.6 atm

Final temperature T2= 250K

Final volume V2=???

From the general gas equation:

P1V1/T1 = P2V2/T2

P1V1T2 = P2V2T1

V2 = P1V1T2/P2T1

V2= 6.7×18×250/5.6×500= 30150/2800

V2= 10.77 L