Question

Suppose 1.000 mol argon (assumed to be an ideal gas) is confined in a strong, rigid container of volume 22.41 L at 273.15 K. The system is heated until 3.000 kJ (3000 J) of heat has been added. The molar heat capacity of the gas does not change during the heating and equals $\mathbf{12}\mathbf{.}\mathbf{47}{\mathbf{JK}}^{\mathbf{-}\mathbf{1}}{\mathbf{mol}}^{\mathbf{-}\mathbf{1}}$.

(a)Calculate the original pressure inside the vessel (in atmospheres).

Short answer

(a)The original pressure inside the vessel is1.000 atm.

Step-by-step solution

The given information.

The number of moles of argon is 1.000 mol.

The volume is 22.41 L

The temperature is 273.15 K.

The amount of heat is 3000 J.

The pressure.

For an ideal gas, the pressure is given by:

$\begin{array}{l}\mathrm{PV}=\mathrm{nRT}\\ \mathrm{P}=\frac{\mathrm{nRT}}{\mathrm{V}}\end{array}$

Calculating the original pressure inside the vessel

To determine the original pressure of argon inside the vessel. The ideal gas equation is given as:

$\mathrm{PV}=\mathrm{nRT}$

P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant and T is the temperature.

Substitute for P in the equation is:

$\begin{array}{c}\mathrm{P}=\frac{\mathrm{nRT}}{\mathrm{V}}\\ =\frac{1.000\mathrm{mol}\times 0.08206{\mathrm{LatmK}}^{-1}{\mathrm{mol}}^{-1}\times 273.15\mathrm{K}}{22.41\mathrm{L}}\\ =1.000\mathrm{atm}\end{array}$

Thus, the original pressure inside the vessel is 1.000 atm.