Gas is confined in a tank at a pressure of 11.0 atm and a temperature of 25.0 degrees C. If two-thirds of the gas is withdrawn and the temperature is raised to 75.0 degrees C, what is the new pressure of the gas remaining in the tank?

Please help!

plz say d answer I dnt knw it

To find the new pressure of the gas remaining in the tank, we can use the combined gas law. The combined gas law states that the initial pressure times the initial volume divided by the initial temperature is equal to the final pressure times the final volume divided by the final temperature. The equation can be written as follows:

(P1 × V1) / T1 = (P2 × V2) / T2

Where:
P1 = initial pressure
V1 = initial volume
T1 = initial temperature
P2 = final pressure (to be determined)
V2 = final volume (since two-thirds of the gas is withdrawn, we can assume the final volume is two-thirds of the initial volume)
T2 = final temperature

Let's plug in the given values:
P1 = 11.0 atm
V1 = initial volume (unknown)
T1 = 25.0 °C = 273.15 K (temperature must be in Kelvin)
P2 = final pressure (to be determined)
V2 = (2/3) × initial volume
T2 = 75.0 °C = 348.15 K (temperature must be in Kelvin)

Now, we need to solve for P2. Rearranging the equation, we get:

P2 = (P1 × V1 × T2) / (V2 × T1)

Since we have an unknown initial volume (V1), we need additional information to find the new pressure.