Question

An open vessel at a temperature of 20 degree celsius is heated at a constant pressure to 340 degree Celsius .what fraction of air originally contained in the vessel has gone out?​

Answer 1

Tthe fraction of air that has gone out is 52.2%.

Explanation:

Since the vessel is open, we can assume the pressure remains constant during the heating process.

Additionally, the volume of the vessel itself remains constant.

Let's denote:

P1: Initial pressure (constant)

V1: Initial volume (constant)

T1: Initial temperature (20°C = 293 K)

n1: Initial number of moles of air

T2: Final temperature (340°C = 613 K)

n2: Final number of moles of air remaining in the vessel

We can use the ideal gas law to relate these variables:

PV = nRT

Since both P and V are constant, we can rewrite the equation:

n1/T1 = n2/T2

Solving for the fraction of air that has gone out:

Fraction gone out = (n1 - n2) / n1 = 1 - n2/n1 = 1 - T1/T2 = 1 - 293 K / 613 K = 1 - 0.478 = 0.522

Therefore, the fraction of air that has gone out is approximately 52.2%.