Question

10) A chemist expands changes from 50 to 150 liters. He finds that 1.55 kJ of heat have been absorbed in the process. Determine the internal energy change that took place. an ideal gas against a constant external pressure of 700 mmHg, and finds its volum

Answer 1

Given data:

• Constant external pressure, P_ex = 700 mmHg
• Initial volume, V₁ = 50 L = 0.050 m³ (Because 1L = 10^-3 m³)
• Final volume, V₂ = 150 L = 0.150 m³
• Amount of heat absorbed, Q = 1.55 kJ = 1550 J

To determine:

• The change in the internal energy.

Solution:

Constant external pressure, P_ex

= 700 mmHg

33.322Ра 700тmHg * 1тm Hg

= 93325.7 Pa

Work done during this process.,

V2 W PerdV

Since the external is maintained at constant level. Therefore, it can be pulled out of the integral.

W Per dV Vi

Integration yields

W -Per(V - V)

Thus the change in internal energy is given by the first law thermodynamics for a closed system.

AUQW

AUQPerAV

AU - Q — Реr (V? — V)

substitute the known values in the above equation, we get

AU (1550J- 93325.7Pa(0.150 0.050)m2

$\Delta$U = - 7782.57 J

or $\Delta$U = - 7.78257 kJ

(Required change in the internal energy. Here negative sign indicates decrease in internal energy).