1litre of an ideal gas (γ = 1.5) at 300 K is suddenly compressed to half its original volume. (a) Find the ratio of the final pressure to the initial pressure. (b) If the original pressure is 100 kPa, find the work done by the gas in the process. (c) What is the change in internal energy? (d) What is the final temperature? (e) The gas is now cooled to 300 K keeping its pressure constant. Calculate the work done during the process. (f) The gas is now expanded isothermally to achieve its original volume of 1 litre. Calculate the work done by the gas. (g) Calculate the total work done in the cycle.

1litre of an ideal gas (γ = 1.5) at 300 K is suddenly compressed to half its original volume. (a) Find the ratio of the final pressure to the initial pressure. (b) If the original pressure is 100 kPa, find the work done by the gas in the process. (c) What is the change in internal energy? (d) What is the final temperature? (e) The gas is now cooled to 300 K keeping its pressure constant. Calculate the work done during the process. (f) The gas is now expanded isothermally to achieve its original volume of 1 litre. Calculate the work done by the gas. (g) Calculate the total work done in the cycle.

Simran Bhatia,
9 years ago

Grade:11

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1 Answers

Kevin Nash

askIITians Faculty
332
Points

9 years ago

Sol. γ = 1.5, T = 300 k, V = 1Lv = 1/2l
(a) The process is adiabatic as it is sudden,
P base 1 V base 1^γ = P base 2 V base 2^γ ⇒ P base 1 (V base 0)^γ = P base 2 (V base 0/2)^γ ⇒ P base 2 = P base 1 (1/1/2)^1.5 = P base 1 (2)^1.5 ⇒ P base 2/P base 1 = 2^1.5 = 2√2
(b) P base 1 = 100 KPa = 10^5 Pa W = nR/γ – 1[T base 1 – T base 2]
T base 1 V base 1^γ-1 = P base 2/V base 2^γ-1 ⇒ 300 * (1)^1.5-1 = T base 2 (0.5)^1.5-1 ⇒ 300 * 1 T base 2 √0.5
T base 2 = 300 * √1/0.5 = 300 √2 K
P base 1 V base 1 = nRT base 1 ⇒ n = P base 1 V base 1/RT base 1 = 10^5 * 10^-3/R * 300 = 1/3R (V in m^3)
W = nR/γ – 1[T base 1 – T base 2] = 1R/3R(1.5 - 1) [300 - 300√2] = 300/3 * 0.5(1 - √2) = -82.8 J = - 82 J.
(c) Internal Energy,
dQ = 0, ⇒ du = – dw = –(–82.8)J = 82.8 J ≈ 82 J.
(d) Final Temp = 300 √2 = 300 × 1.414 × 100 = 424.2 k ≈ 424 k.
(e) The pressure is kept constant. ∴ The process is isobaric.
Work done = nRdT = 1/3R * R * (300 – 300 √2) Final Temp = 300 K
= - 1/3 * 300 (0.414 = - 41.4 J. Initial Temp = 300 √2
(f) Initial volume ⇒ V base 1/T base 1 = V base 1’/T base 1’ = V base 1’ = V base 1/T base 1 * T’ base 1 = 1/2 * 300 * √2 * 300 = 1/2√2 L.
Final volume = 1L
Work done in isothermal = nRT In V base 2/V base 1
= 1/3R * R * 300 In(1/1/2√2) = 100 * In (2√2) = 100 * 1.039 = 103
(g) Net work done = W base A + W base B + W base C = - 82 – 41.4 + 103 = - 20.4 J.

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