A rigid vessel of fixed volume contains an ideal gas at an initial pressure of 1.50 \times… — JAMB UTME

A rigid vessel of fixed volume contains an ideal gas at an initial pressure of $1.50 \times 10^5\text{ Pa}$ and a temperature of $27^\circ\text{C}$ . Additional gas is pumped into the vessel until the total number of moles of gas is doubled. If the temperature of the gas increases to $87^\circ\text{C}$ , what is the final pressure of the gas in units of $10^5\text{ Pa}$ ?

Answer: 3.6 10^5 Pa

Answer

The final pressure of the gas is

3.6 \times 10^5\text{ Pa}

(which is

3.6

in units of

10^5\text{ Pa}

Using the ideal gas equation

PV = nRT

at fixed volume, the ratio of final to initial pressure is given by

P_2/P_1 = (n_2/n_1) \times (T_2/T_1)

. Converting temperatures to Kelvin yields

T_1 = 300\text{ K}

and

T_2 = 360\text{ K}

. Given that the number of moles doubles (

n_2/n_1 = 2

), substituting the values gives

P_2 = 1.50 \times 10^5 \times 2 \times (360/300) = 3.60 \times 10^5\text{ Pa}

Step-by-Step Solution

Convert initial and final temperatures from Celsius to absolute temperature in Kelvin.

T_1 = 27 + 273 = 300\text{ K}

and

T_2 = 87 + 273 = 360\text{ K}

Gas laws require absolute temperatures in Kelvin for thermodynamic calculations.

Formulate the pressure relation from the ideal gas equation

PV = nRT

Since volume

V

and the universal gas constant

R

are constant,

P_2 = P_1 \times \frac{n_2}{n_1} \times \frac{T_2}{T_1}

Pressure is directly proportional to both the number of moles and the absolute temperature when volume is fixed.

Substitute the mole ratio

n_2/n_1 = 2

, initial pressure, and Kelvin temperatures to compute

P_2

P_2 = 1.50 \times 10^5 \times 2 \times \frac{360}{300} = 3.60 \times 10^5\text{ Pa}

Evaluates the final gas pressure in the requested numerical units.

Key Concept

Ideal Gas Equation and Variable Moles under Constant Volume

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