Thank you for registering.

One of our academic counsellors will contact you within 1 working day.

Please check your email for login details.
MY CART (5)

Use Coupon: CART20 and get 20% off on all online Study Material

ITEM
DETAILS
MRP
DISCOUNT
FINAL PRICE
Total Price: Rs.

There are no items in this cart.
Continue Shopping

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.

Grade:11

1 Answers

Kevin Nash
askIITians Faculty 332 Points
7 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.

Think You Can Provide A Better Answer ?

Provide a better Answer & Earn Cool Goodies See our forum point policy

ASK QUESTION

Get your questions answered by the expert for free