Heat is added to two identical samples of a monatomic ideal gas. In the first sa

Question

Heat is added to two identical samples of a monatomic ideal gas. In the first sample the heat is added while the volume of the gas is kept constant, and the heat causes the temperature to rise by 87 K. In the second sample, an identical amount of heat is added while the pressure (but not the volume) of the gas is kept constant. By how much does the temperature of this sample increase?

Explanation / Answer

1st > Q heat is added constant Volume > dV =0 work done by gas = dW = pdV = 0 dQ = dU + dW Q = n cv dT = 47n *cv ------------- (1) ------------------------------ 2nd> constant P dW = P[dV] = p[V2 - V1] gas equation> PV1 = nRT1, PV2 = n R T2 P(V2-V1) = dW = n R [delta T] dQ = dU + dW Q = n cv delta T + n R [delta T] = n[cv + R] delta T Q = n cp delta T 47 n cv = n cp delta T delta T = 47 cv/cp for monoatomic gas> cv = 3R/2, cp = 5R/2 delta T = 47[3/5] = 2.8 K obviously delta T here is less because some heat goes to work against surroundings in the1st, heat has sole job of increasing the temp.

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