CALC The temperature of 0.150 mol of an ideal gas is held constant at 77.0o C while its volume is reduced to 25.0% of its initial volume. The initial pressure of the gas is 1.25 atm. (a) Determine the work done by the gas. (b) What is the change in its internal energy? (c) Does the gas exchange heat with its surroundings? If so, how much? Does the gas absorb or liberate heat?
Solution 27E Step 1 of 5: 0 In the given process, temperature remains constant at T=77 c. Since an isothermal process is a constant-temperature process. For a process to be isothermal, any heat flow into or out of the system must occur slowly enough that thermal equilibrium is maintained. Step 2 of 5: (b) What is the change in its internal energy In general; in case of isothermal, none of the quantities like change in internal energy( U ), work done(W) and energy supplied (Q) is zero.But in some special cases like ideal gas,where the internal energy of a system depends only on its temperature, not on its pressure or volume. For such systems like ideal gas, if the temperature is constant, the internal energy is also constant; and That is, any energy entering the system as heat Q must leave it again as work W done by the system. Hence U = 0 W= Q Therefore, change in internal energy is zero in the given case. Step 3 of 5: (a) Determine the work done by the gas. Given data, Number of moles, n=0.150 mol Gas constant , R= 8.314 J/mol.K Temperature, T = T = T = 77 c 0 1 2 Using T = T + c73.15 T = 77 + 273.15 K T = 350.15 K Initial volume, V = V 1 Final volume, V =20.25 V To find, Work done, W=