Question

Question: The PV plot (Figure 1) depicts 154 mols of a gas going from state A with PA = 1.96 X 10 Pa and VA = 1.41 m to state B with Figure 1: A PV plot depicting the described gas going up and right from state A to state B via a straight diagonal pathway. The x-axis represents volume and the y-axls represents pressure. Part 1) How much work is done on the gas as it goes from state A to state B?

Answer 1

Part 1 )

P_A = 1.96 * 10⁵ Pa

V_A = 1.41 m³

P_B = 3.43 * 10⁵ Pa

V_B = 3.35 m³

Work done is given by area under P-V curve,

SInce area under Curve between A and B is a trapezium,

W_A->B = (1/2) * (V_B - V_A) * (P_A + P_B ) = 5.2283 * 10⁵ J (Since the gas is expanding positive work is done by the system)

Part 2 )

Change in internal energy is given by, $\Delta$U = nC_v$\Delta$T

where, n = no. of moles ; C_v = specific heat capacity at constant volume ; $\Delta$T = change in temperature from A to B

$\Delta$​​​​​​​T = (P_BV_B / nR) - (P_AV_A / nR) = (1 / nR)(P_BV_B -P_AV_A) = (1/nR) * 8.7269 * 10⁵

Let degrees of freedom be f, therefore C_v = (f/2) * R

1.31 * 10⁶ = (f/2) * 8.7269 * 10⁵

f = 3

Part 3 )

Heat added,Q = $\Delta$U + W = 1.31 * 10⁶ + 5.2283 * 10⁵ = 1.83283 * 10⁶ J