50.0 moles of an ideal gas for which Cv = 5/2R and Cp = 7/2R initially has a pre

Question

50.0 moles of an ideal gas for which Cv = 5/2R and Cp = 7/2R initially has a pressure of 1.0 × 10^5P a, and a volume of 1.0m^3 . It undergoes a process where the pressure varies with volume according to p(V ) = 0.5 × 10^5 (Pa) + 2 × 105^(Pa/m^3) V 0.5 × 10^5 (Pa/m^6)V^2 (1) as it expands to a final volume of 3.0m3 .

(a). What is the change in temperature for this process? (a) 240K (b) ** 480K (c) 720K (d) 960K (e) 1200K

(b). What is the work done by the gas during this process? (a) 1.0 × 10^5J (b) ** 2.7 × 10^5J (c) 4.3 × 10^5J (d) 5.3 × 10^5J (e) 8.0 × 10^5J

(c). What is the change in internal energy of the gas during this process (recall that the change in internal energy does not depend on path taken between states) (a) 2.5 × 10^5J (b) 3.0 × 10$5J (c) ** 5.0 × 105J (d) 7.0 × 10^5J (e) 7.5 × 10^5J

(d) . How much energy enters the gas as heat during this process? (a) 5.7 × 10^5J (b) 7.3 × 10^5J (c) ** 7.7 × 10^5J (d) 8.5 × 10^5J (e) 8.8 × 10^5J

Explanation / Answer

P(V=3)=-0.5*105 + 2*105*3 - 0.5*105*32=1*105 pressure is same at both point

Ti =P1v1/nR=1*105*1/50*8.314=240.5580K

Tf=P2V2/nR=1*105*3/50*8.314=721.674K

temperature differance=481.116

b is the answere

part b

work done by gas during process=nR*temp differance=50*8.314*481.116=199999.92J=200KJ arround

partc

U=nCvdT=50*5/2*8.314*481.116=500KJ =5*105J

part d

Q=nCpdT=50*7/2*8.314*481.116=6.99999*105J

in ans b and d have wronw value it should be 2*105 and we know

Q=U+W so it should be 7*105

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