Question

Using 0.02 mol of an ideal monoatomic gas, an isochoric process from A (230 kPa, 1 L) to B (98 kPa, 1 L) results in what change in internal energy

Answer 1

n = 0.02 mol

V = 1 L = 0.001 m³ = constant

$P_1 = 230 \times 10^3 \;Pa$

$P_2 = 98 \times 10^3 \;Pa$

change in internal energy $∆U = nC_V(T_2-T_1)$

For an ideal monoatomic gas $C_V = \frac{3R}{2}$

$∆U = n(\frac{3R}{2})(\frac{P_2V}{nR}-\frac{P_1V}{nR})$

$∆U = \frac{3}{2}(P_2V -P_1V)$

$∆U = \frac{3}{2}(98 \times 10^3 \times 0.0001 - 230 \times 10^3 \times 0.001)$

$∆U = 1.5 \times (98 – 230) = -198 \;J ≈ -200 \;J$