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

4.24 L

Step-by-step 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;

`(P1V1)/(T1)=(P2V2)/(T2)`

Rearranging the formula;

`V2=(P1V1T2)/(P2T1)`

Therefore;

`V2=((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