We have a container of 1.14 moles of an ideal monatomic gas. Thevolume of the co

Question

We have a container of 1.14 moles of an ideal monatomic gas. Thevolume of

the container is 15.0 liters, and the temperature of the gas is21.7?C. We compress the

gas adiabatically to 13.8 liters. (a) Find the final temperature(K) of the gas.

Neglect any heat ?ow into the surroundings. Caution: Be sure to usethe ideal value

for , not the approximate value. (b) Find the change ininternal energy (J) of the gas.

(c) Find the work done (J) on the gas. Be sure to include thecorrect signs on the

answers.

Explanation / Answer

a) P1*V1 = n*R*T1 P1 = n*R*T1/V1 = (1.14 mol)*(0.0821 L*atm/mol/K)*(294.7K)/(15L) =1.83881012 atm P1*V1 =P2*V2         = 5/3; this is the adiabatic process equation so P2/P1 =(V1/V2) P1*V1/T1 = P2*V2/T2 T2 = (P2/P1)*(V2/V1)*T1 T2 = (V1/V2) *(V2/V1)*T1 = (V1/V2)5/3*(V2/V1)*T1 T2 = (V1/V2)2/3 *T1 = (15/13.8)2/3*(294.7K) = 311.545573K b) delU = Cv*delT =Cv*(T2-T1),             Cv = (3nR/2)     where R = 8.314 J/mol/K delU = (3*1.14 mol *8.314 J/mol/K /2) *(311.545573 K -294.7K) = 239.5 joules c) delU = Q + W but the process is adiabatic, so Q = 0 Thus delU = W = 239.5 joules This is the work done on the system PM me if you have questions

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