Comparing Thermodynamic Processes. In a cylinder, 1.20 mol of an ideal monatomic gas, initially at 3.60 X 105 Pa and 300 K, expands until its volume triples. Compute the work done by the gas if the expansion is (a) isothermal; (b) adiabatic; (c) isobaric. (d) Show each process in a pV-diagram. In which case is the absolute value of the work done by the gas greatest? Least? (e) In which case is the absolute value of the heat transfer greatest? Least? (f) In which case is the absolute value of the change in internal energy of the gas greatest? Least?

Solution 68P a.) An isothermal process is a constant-temperature process. Step 1 of 2: Work done by an ideal monatomic gas in expanding isothermally from a volume of (V) to three times its volume(3V) initially at a pressure of 3.6 × 10 5 Pa and temperature of 300k can be computed as vf W = P × dV vi = n × R × T ×ln V [/V i ] = 1.20× 8.31 × 300 ×ln 3V[/V ] = 1.20 × 8.31 × 300 × ln 3 [ ] = 1.20× 8.31 × 300 × 1.098 W = 3286.6 J Step 2 of 2: For an ideal gas the internal energy is directly proportional to temperature, it follows that there is no change in internal energy of the gas during an isothermal process . The First law of thermodynamics then becomes U = 0 = Q W Q = W All the heat added to the system is used to do work. The value of heat transfer is equal to the amount of work done by the gas. Q = W = 3286.6 J Step 3 of 3: Since the gas is expanding isothermally,there is no change in the internal energy of the gas. U= 0 b.) An adiabatic process is defined as one with no heat transfer into or out of a system; Q=0. The First law of thermodynamics then becomes Q = U + W 0 = U + W U = W Work done by an ideal monatomic gas in expanding adiabatically from a volume of (V) to three times its volume(3V) initially at a pressure of 3.6 × 105 Pa and temperature of 300k can be computed as W = n × C × vT Where, n = number of moles of monatomic gas C v Molar heat capacity at constant volume dT = Change in temperature of the gas when it expands adiabatically (T T ) 1 2 Step 1 of 1: For an adiabatic process; 1 T × V = constant T 1 V 11 = T 2 V 21 T = T × (V /V ) 1 2 1 1 2 1.671 = 300 × (V /3 × V ) 0.67 = 300 × (1/3) = 300 × 0.478 T = 1 43.4 K 2 Step 2 of 2: Now work done by the gas can be computed as follows; W = n × C × vT = n × C ×v( T 1 ) 2 = 1.2 × 12.47 × (300 143.4) W = 2 343.36 J Step 3 of 3: Since the gas is expanding adiabatically ,there is no transfer of heat into or out of the system;Q= 0. First law of thermodynamics then becomes U = W The value of the heat transfer(Q) is equal to zero. Q = 0 The value of change in the internal energy of the gas is given by U = W = 2343.36 J c.) An isobaric process is a constant-pressure process. In general, none of the three quantities U,Q, and W is zero in an isobaric process. Work done by an ideal monatomic gas in expanding isobarically from a volume of (V) to three times its volume(3V) initially at a pressure of 3.6 × 105 Pa and temperature of 300k can be computed as Step 1 of 1: W= P × dV = n × R × dT Where, n = number of moles of monatomic gas R = universal gas constant dT = T 2 T 1 Step 2 of 2: For an isobaric process; V 1 / V 2 = T 1 T 2 T 2 = (V /2V ) 1 T 1 = (3V / V ) × 300 = 3 × 300 T = 9 00 K 2 Step 3 of 3: Now, work done by the gas can be computed as W = n × R × (T 2 T 1 = 1.2 × 8.31 × (900 300) = 1.2 × 8.31 × 600 W = 5 983.2 J