Two rigid vessels A and B , each having a volume of 200 cm^3 , contain an ideal gas ( C_V = 12.5 J K^ -1 mol^ -1 ). These vessels are connected to a manometer tube containing mercury. Initially, the pressure in both

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Quantrex Academy · Accepted Answer

Correct answer: C. 12.5 cm

For a constant volume process, the heat supplied is

Q = n C_V \Delta T

.

From the ideal gas equation

PV = nRT

, the change in pressure at constant volume is

\Delta P = \dfrac{n R \Delta T}{V}

.

Substituting

n \Delta T = \dfrac{Q}{C_V}

, we get

\Delta P = \dfrac{R Q}{V C_V}

.

The difference in pressure between vessel B and vessel A is:

\Delta P_{net} = \Delta P_B - \Delta P_A = \dfrac{R}{V C_V} (Q_B - Q_A)

Substituting the given values (

R = 8.3 \text{ J K}^{-1} \text{ mol}^{-1}

,

V = 200 \times 10^{-6} \text{ m}^3

,

C_V = 12.5 \text{ J K}^{-1} \text{ mol}^{-1}

,

Q_B - Q_A = 5.0 \text{ J}

):

\Delta P_{net} = \dfrac{8.3 \times 5.0}{2 \times 10^{-4} \times 12.5} = 16600 \text{ N m}^{-2}

To convert this pressure difference into cm of Hg, we use

\rho_{Hg} = 13600 \text{ kg m}^{-3}

and

g = 9.8 \text{ m s}^{-2}

:

1 \text{ cm of Hg} = 0.01 \times 13600 \times 9.8 = 1332.8 \text{ N m}^{-2}

The difference in height

h

is:

h = \dfrac{16600}{1332.8} \approx 12.45 \text{ cm}

This value is approximately

12.5 \text{ cm}

.

Answer:

12.5 \text{ cm}

Two rigid vessels A and B , each having a volume of 200 cm^3 ,...

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