Question

A gas weighing 12.5g is contained in a constant volume vessel at 27•C and a certain pressure.The temperature is now increased to 352•C. Calculate mass of the gas released from the vessel if pressure remains constant.​

Answer 1

Answer:

Explanation:

The temperature is now increased to 352•C. Calculate mass of - 1201 ... A gas weighing 12.5g is contained in a constant volume vessel at 27•C and a certain ... of the gas released from the vessel if pressure remains constant.

Answer 2

Given:

A gas weighing 12.5g is contained in a constant volume vessel at 27°C and a certain pressure.

The temperature is now increased to 352°C.

To find:

The mass of the gas released from the vessel if the pressure remains constant.​

Solution:

As per the Ideal Gas Law, we have

$\boxed{\bold{PV = nRT}}$

where P = pressure, V = Volume, n = no. of moles, R = Ideal gas constant and T = temperature

The volume and pressure inside the vessel remain constant in both cases.

Initial condition:

Mass of the gas, m₁ = 12.5 g

The temperature of the vessel, T₁ = 27°C = 27 + 273 K = 300 K

Therefore, using the Ideal Gas Law, we get

$PV =n_1 RT_1$

$\implies PV =\frac{m_1}{M_1} RT_1$

$\implies PV =\frac{12.5}{M_1} R\times 300$ Equation1

Final condition:

Let " m₂" be the mass of the released gas.

The temperature of the vessel, T₂ = 352°C = 352 + 273 K = 625 K

Therefore, using the Ideal Gas Law, we get

$PV =n_2 RT_2$

$\implies PV =\frac{m_2}{M_2} RT_2$

$\implies PV =\frac{m_2}{M_2} R\times 625$ Equation2

Comparing equation1 and equation2, we get

$\frac{12.5}{M_1} \times 300 = \frac{m_2}{M_2} \times 625$

the molar mass of the gas will be the same ∴ M₁ = M₂

$\implies 12.5 \times 300 =m_2 \times 625$

$\implies m_2 = \frac{ 12.5 \times 300 }{625}$

$\implies m_2 = \frac{ 3750 }{625}$

$\implies \bold{m_2 = 6\:g}$

Thus, the mass of the gas released from the vessel if the pressure remains constant will be → 6 g.

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