InterviewSolution
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InterviewSolution
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One mole of a gas is enclosed in a cylinde in a cyclinder and occupies a volume of `1.5 L` at a pressure 1.5 atm. It is subjected to strong heating due to which temperature of the gas increase according to the relation `T = alpha V^(2)`, where `alpha` is a positive constant and `V` is volume of the gas. a. Find the work done by air in increasing the volume of gas to `9 L`. b. Draw the `P - V` diagram of the process. c. Determine the heat supplied to the gas (assuming `gamma = 1.5)`. |
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Answer» `W = int_(V_(1))^(V_(2)) PdV = int_(V_(1))^(V_(2)) (RT)/(V) dV = R int_(V_(1))^(V_(2)) (alpha V^(2))/(V) dV` `alpha R int_(V_(1))^(V_(2)) VdV = (alphaR)/(2) (V_(2)^(2) - V_(1)^(2))``alpha R int_(V_(1))^(V_(2)) VdV = (alphaR)/(2) (V_(2)^(2) - V_(1)^(2))` `W = (R )/(2) (T_(2) - T_(1))` (i) Now, `T_(1) = alpha V_(1)^(2)` `T_(2) = alpha V_(2)^(2)` `T_(2) = T_(1) [(V_(2))/(V_(1))]^(2) = 35 T_(1)` Hence from Eq. (i) we have `W = (R )/(2) [36 T_(1) - T_(2)] = (35 RT_(1))/(2) = (35)/(2) P_(1) V_(1)` `= (35)/(2) xx (1.5 xx 10^(-3)) xx (1.2 xx 10^(5)) J = 31 50 J` b. We know, `PV = RT = R alpha V^(2)` `P prop V` Hence `P -V` graph is a straight line. c. `Delta U = n V_(V) Delta T = n ((R )/(gamma - 1)) (T_(2) - T_(1))` `(R )/(0.5) xx 35 T_(1) = 70 RT_(1) = 70 P_(1) V_(1)` `= 70 xx (1.2 xx 10^(5)) xx (1.5 xx 10^(-3)) J = 12600 J` `Delta Q = Delta U + W 12600 + 3150 = 15750 J` |
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