Question

Gas is confined in a tank at a pressure of 1.06E+6 Pa and a temperature of 14.3

Answer 1

Initial Condition:

P₁ = 1.06E+6 Pa, T₁ = 14.3^oC = 14.3+273 = 287.3K

Say volume is V₁ and number of moles of gas is n₁

Final condition:

P₂ = ?, T₂ = 61.6^oC = 334.6K

V₁ = V₂ (Since tank remains the same)

$n_{2} = \frac{n_{1}}{2}$ (Since half of the gas is removed)

On using ideal gas equations for both the conditions and diving them, we get

$\frac{P_{2}V_{2}}{P_{1}V_{1}}=\frac{n_{2}RT_{2}}{n_{1}RT_{1}}$

$\Rightarrow \frac{P_{2}}{1.06\times 10^{6}}=\frac{334.6}{2\times287.3}$

$\Rightarrow P_{2} = 1.06\times 10^{6}\times \frac{334.6}{2\times287.3} = 6.17\times 10^{5} Pa}$

which is the new pressure in the tank.