Question

The pressure, volume and temperature of a mole of an ideal gas are related by the equation \(PV = 8.317\), where \(P\) is measured in kilopascals, \(V\) in liters and \(T\) in kelvins. Use differentials to find the approximate change in the pressure if the volume increases from \(12\,\,L\) to \(12.3\,\,L\) and the temperature decreases from \(310K\) to \(315K\).

Short answer

The approximate change in pressure is in negative that is pressure decreases and pressure decreases by \(8.829\,kpa\).

Step-by-step solution

Given and To Find.

\(PV = 8.317\)

Volume increases from\(12\,\,L\)to\(12.3\,\,L\)

Temperature decreases from\(310k\)to\(305\;K\)

The approximate change in the pressure (To find)

Finding \(p\).

\(PV = 8.31T\)

\(V = 12\)

\(T = 310\)

\(P \times 12 = 3.31 \times 310\)

\(12P = 2,576.1\)

\(P = \frac{{2576.1}}{{12}}\)

\(P = 214.675\)

Find change in volume and temperature.

\(PdV + VdP = 8.31dT\)

\( dV = 12.3 - 12\)

\( = 0.3\)

\(\begin{aligned}{l}dT = 305 - 310\\ = - 5\\v = 12\\p = 214.675\end{aligned}\)

Find Change in Pressure.

\(PdV + VdP = 8.31dT\)

\(214.675 \times 0.3 + 12 \times dP = 8.31 \times ( - 5)\)

\(64.4025 + 12dP = - 41.55\)

\(12dP = - 41.55 - 64.40255\)

\(12dP = - 105.9525\)

\(dP = \frac{{ - 105.9525}}{{12}}\)

\(dP = - 8.829375\)

Therefore, The change in pressure is \( - 8.829375\). So, The pressure decreases by \( - 8.829 kpa\)