Question

Show work for partial credit. Assume 3 significant figures for all numerical values. Write all answers on the sheets provided; work written on the test sheet will not be graded. g= 9.80 m/s 1 atm = 101 kPa 1L = 10 m G-6.67 x 10''Nm²/kg? 1. (12 points) Compute the heat absorbed by and the work done on 0.400 moles of a diatomic gas (Cy = 2.5R, Cp = 3.5R ) for each of the following processes (give your final answers in Joules): a. The gas is compressed from 5.00 L to 2.00 L while the pressure is held constant at 2.00 atm. b. The pressure is increased from 1.00 atm to 3.00 atm in a container of volume 6.50 L.

Answer 1

Ans:- (a) No of moles of a diatomic gas n = 0.4 moles. a = 2.5R p= 3.5R since gas is compressed at constant pressure P= 2.0 atm, = 2 x 1.01 Xlos Pa = 2:02x1.5 pa initial volume = 5 L = 5x10-3 final volume V2 = 2 L = 2 x 10-3 m? Hence the work done on the system is - W = P (₂-v) =- 2.02 x 103 x 3x10 3 W = – 6.06 X 102 I Heat absorbed in Cp = 1 (do this process is - dono dT ndT Tp Hence heat absorbed by the system is - A9 = NATI from ideal gas equation. PV = nRT Differentiating the above equation - PAV = nROT NAT = PAY = W. NAT = – 0.73x10² molk AQ- 3.5 R X W = 3.5w. AQ = – 3.5 X 6.06 4102 – – 21.21 X 102 I I AQ = -2 12 x 3 J

(b) the pressure is increased at constant volume V = 6.5L = 6.5x 10-3 m3 Initial pressure is P. = I atm = 1.01 X 10 pa Final pressure is P2 = 3 atm. = 3.03 × 10 pa since this is an isochoric process and the work done in an isochoric process is zero. Hence W = o A9 = nCAT PV = nRTI P₂ V = nRT2 - 3 = T- 3T, и к PV 101x10² x 6.5 x 103 0.4 x 8.31 T = 1.975 x 16²k T2 = 3T, = 3x 1.975 x 10² k = 5.925 x 10²k AT = T2-T, = (5.925-1.975) x 10² k AT= 3.95 x 10²k Heat absorbed is - AQ = 0.4x 2.588.31 x 3.958102 D9 = 3.28 x 103 J