I need help for this question in combining gas law P1V1/T1=P2V2/T2.

P1V1/T1=P2V2/T2


P1 is 1.5 atm

V1 is a question mark.

T1 is -20 °C K

P2 is 2.6 atm

V2 is 750 mL with a blank box underneath it.

T2 is -5 °C K.

How can I combine this question?

P1V1/T1=P2V2/T2

I'm not sure you copied all of the problem. I don't know what you're combining. You are making some conversions and some substitutions. Here is some help on what I can determine from your numbers.

P1 is 1.5 atm

V1 is a question mark.
I suppose the question mark means that you are solving for V1 and I assume it is to be determined in liters.

T1 is -20 °C K
What you have written doesn't make sense to me. I assume T1 is -20 C and you want to convert that to Kelvin. To convert to Kelvin, remember Kelvin = 273 + C.

P2 is 2.6 atm
Nothing needs to be done to this since P1 is in atm, also.

V2 is 750 mL with a blank box underneath it.
My best guess is that 750 mL is to be converted to liters. That is done by using the conversion factor of 1 L = 1000 mL.
750 mL x (1 L/1000 mL) = ?? Liters.


T2 is -5 °C K.
Again, I must assume T2 is -5 C and you want to convert to Kelvin. See above conversion for T1.

Plug all those numbers into the formula and solve for V1. The answer will be in liters.

To solve this problem, you can use the combined gas law formula, which is P1V1/T1 = P2V2/T2. This formula relates the pressure, volume, and temperature of a gas between two different states.

Given values:
P1 = 1.5 atm
V1 = ?
T1 = -20 °C = -20 + 273.15 K = 253.15 K
P2 = 2.6 atm
V2 = 750 mL = 750/1000 L = 0.75 L
T2 = -5 °C = -5 + 273.15 K = 268.15 K

Since we are solving for V1, we need to rearrange the formula to solve for V1.

P1V1/T1 = P2V2/T2
V1 = (P2 * V2 * T1) / (P1 * T2)

Now, let's substitute the given values into the formula and calculate V1:

V1 = (2.6 atm * 0.75 L * 253.15 K) / (1.5 atm * 268.15 K)

By multiplying the values inside the brackets, then dividing the products in a separate step, you can compute the result:

V1 ≈ 0.777 L

Therefore, the value of V1 is approximately 0.777 L.