If a gas occupies 30 l at 1.3 atm and 300 k what volume will it occupy at stp

Let's assume that the given gas is an ideal gas.

We can use combined gas law,

PV/T = k (constant)

Where, P is the pressure of the gas, V is volume of the gas and T is the temperature of the gas in Kelvin.

For two situations, we can use that as,

P₁V₁/T₁= P₂V₂/T₂

P₁ = 1.3 atm

V₁ = 30 L

T₁ = 300 K

P₂ = Standard pressure = 1 atm

V₂ = ?

T₂ = Standard temperature = 273 K

By applying the formula,

    1.3 atm x 30 L / 300 K = 1 atm x V₂ / 273 K

                                   V₂ = (1.3 atm x 30 L x 273 K) / (300 K x 1 atm)

                                   V₂ = 35.49 L

Hence, the volume of the gas at STP is 35.49 L.

Answer : Using t he ideal gas equation, which is;

PV=nRT.

Where, P = 1.3 atm, T = 300K.

We can consider, n and R are constant.

now, on comparing the gas at STP conditions (At standard pressure and temperature),

P′ = 1 atm, and T′ = 273K;

V′ = ?

So, the ideal gas equation is modified as P′V′=nRT′.

Dividing the equations,P′/P X V′/V=T′/T

Therefore, (1 atm / 1.3 atm) X (V' / 30 L) = (273 K / 300 K)

V' = ~ 20.9 L

So, the volume of gas at STP will be ~ 20.9 L

c, liquids evaporate, creating a gas