Question

A sample of an ideal gas has a volume of 3.10 L at 13.80°C and 1.30atm. What is the volume of the gas at 25.00°C and 0.988atm

Answer 1

Answer:

4.24 L

Explanation:

We are given;

• Initial Volume, V1 is 3.10 L
• Initial temperature, T1 is 13.80°C

But, K = °C + 273.15

• Thus, Initial temperature, T1 is 286.95 K
• Initial pressure, P1 is 1.30 atm
• New temperature, T2 is 25°C or 298.15 K
• New pressure, P2 is 0.988 atm

We are required to calculate the new volume;

• We are going to use the combined gas law;

• According to the combined gas law;

$\frac{P1V1}{T1}=\frac{P2V2}{T2}$

Rearranging the formula;

$V2=\frac{P1V1T2}{P2T1}$

Therefore;

$V2=\frac{(1.30atm)(3.10L)(298.15K)}{(0.988atm)(286.95)}$

$V2=4.238L$

$V2=4.24L$

Therefore, the new volume of the gas sample is 4.24 L