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# An ideal gas initially at 335 K undergoes an isobaric expansion at 2.50 kPa. ?

An ideal gas initially at 335 K undergoes an isobaric expansion at 2.50 kPa. The volume increases from 1.00 m3 to 3.00 m3 and 10.6 kJ is transferred to the gas by heat.

(a) What is the change in internal energy of the gas?

(b) What is the final temperature of the gas?

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• An ideal gas initially at 300 K undergoes an isobaric expansion at 2.50 kPa.

(a) If the volume increases from 1.00 m3 to 3.00 m3 and if 15.0 kJ is transferred to the gas by heat, what is the change in its internal energy?

(b) What is its final temperature?

(a)

Change of Internal energy equals heat transferred to plus work done on the gas

ΔU = Q + W

Work done on the gas by changing Volume from V₁ to V₂ is given by the integral

W = - ∫ p dV from V₁ to V₂

for isobaric process ( p=constant)

W = - p·(V₂ - V₁)

for this problem

W = - 2.5×10^3Pa · (3m³ - 1m³) = -5kJ

Hence

ΔU = 10.6kJ - 5kJ = 5.6kJ

(b)

Use ideal gas law

p·V = N·R·T

Since pressure p and N stay constant throughout the process:

V·T = p / (N·R) = constant

hence:

V₁·T₁ = V₂·T₂

<=>

T₂ = T₁· (V₁/V₂)

= 335K · (1m³ / 3m³)

= 111.67K

Source(s): schmiso