Given 6193 ml of a gas at 62.3 °c. what is its volume at 38.1 °c?

5746.0 mL.

Explanation:

We can use the general law of ideal gas: PV = nRT.

where, P is the pressure of the gas in atm.

V is the volume of the gas in L.

n is the no. of moles of the gas in mol.

R is the general gas constant,

T is the temperature of the gas in K.

If n and P are constant, and have two different values of V and T:

V₁T₂ = V₂T₁

V₁ = 6193.0 mL, T₁ = 62.3°C + 273 = 335.3 K.

V₂ = ??? mL, T₂ = 38.1°C + 273 = 311.1 K.

∴ V₂ = V₁T₂/T₁ = (6193.0 mL)(311.1 K)/(335.3 K) = 5746.0 mL.

2b2t

explanation:

2b2t