IF A SAMPLE OF GASE HAVE A VOLUME OF 8.20 L AT 25*C AND 2.00 ATM, HOW MUCH VOLUME WILL IT TAKE UP IF YOU DECREASE THE PRESSURE TO 1.50 ATM AND INCREASE THE TEMPERATURE TO 100*C?

(P1V1/T1) = (P2V2/T2)

Remember T must be in kelvin.

yes, by adding 273.15 right?

yes, STP T is 273.15. 100 C is 373.15 K.

10.9 L

To answer this question, we can use the combined gas law formula, which relates the pressure, volume, and temperature of gases assuming the amount of gas and the gas constant remain constant. The formula is expressed as:

(P1 * V1) / (T1) = (P2 * V2) / (T2)

Where:
P1 and P2 are the initial and final pressures,
V1 and V2 are the initial and final volumes,
T1 and T2 are the initial and final temperatures.

Given:
P1 = 2.00 atm
V1 = 8.20 L
T1 = 25°C (298 K)
P2 = 1.50 atm
T2 = 100°C (373 K)

Now, let's plug in the given values and solve for V2:

(2.00 atm * 8.20 L) / (298 K) = (1.50 atm * V2) / (373 K)

Simplifying the equation:

16.40 atm*L / 298 K = (1.50 atm * V2) / 373 K

To find V2, we can rearrange the equation:

V2 = (16.40 atm*L * 373 K) / (298 K * 1.50 atm)

Calculating the result:

V2 = 20.57 L

Therefore, the volume of the gas sample would be approximately 20.57 L when the pressure is decreased to 1.50 atm and the temperature is increased to 100°C.