Question

A gas at 110 kPa and 30.0 0C fills a flexible container with an initial volume of 2.00L. If the temperature is raised to 80.0 0C and the pressure increases to 440 kPa, what is the new volume? SHOW YOUR WORK FOR FULL CREDIT

Answer 1

To solve this problem, we can use the combined gas law, which relates the pressure, volume, and temperature of a gas:

(P1V1)/T1 = (P2V2)/T2

where P1, V1, and T1 are the initial pressure, volume, and temperature of the gas, and P2, V2, and T2 are the new pressure, volume, and temperature of the gas.

We are given:

P1 = 110 kPa

V1 = 2.00 L

T1 = 30.0°C = 303.15 K

We need to find V2, given:

P2 = 440 kPa

T2 = 80.0°C = 353.15 K

Substituting these values into the combined gas law equation, we get:

(110 kPa)(2.00 L)/(303.15 K) = (440 kPa)(V2)/(353.15 K)

Solving for V2, we get:

V2 = (110 kPa)(2.00 L)(353.15 K)/(303.15 K)(440 kPa) = 1.55 L

Therefore, the new volume of the gas is 1.55 L.