. We have a container of 1.99 moles of an ideal monatomic gas. The volume of the

Question

. We have a container of 1.99 moles of an ideal monatomic gas. The volume of the container is 15.0 liters, and the temperature of the gas is 21.7C. We compress the gas adiabatically to 13.6 liters. (a) Find the final temperature (K) of the gas. Neglect any heat flow into the surroundings. Caution: Be sure to use the ideal value for (the fraction,) not the approximate (decimal) value. (b) Find the change in internal energy (J) of the gas. (c) Find the work done (J) on the gas. Be sure to include the correct signs on the answers

Explanation / Answer

T2 = T1 (V1 / V2)y-1

so T2 = 294.7 (15/13.6)0.66

T2 = 314.387 K

.e T2 = 41.387 celcius

for 1.99 moles of ideal monoatomic gas with given volume and initial temperature... Pi = nRT/Vi = 325.07 Kpa.... So pv^y = K = 298.45

so work done = K (Vf^(1-y) - Vi ^(1-y)) / (1-y) = - 0.4937 KJ = - 493.7 J

as it is adiabatic process. change in internal energy is through the work done..

so change in internal energy = 493.7J

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