A sample of gas, initially with a volume of 1.0 L, undergoes a thermodynamic cycle. Find the work done by the gas on its environment during each stage of the cycle described below. (Enter your answers in J.)(a)First, the gas expands from a volume of 1.0 L to 6.5 L at a constant pressure of 6.0 atm. J(b)Second, the gas is cooled at constant volume until the pressure falls to 1.5 atm. J(c)Third, the gas is compressed at a constant pressure of 1.5 atm from a volume of 6.5 L to 1.0 L. (Note: Be careful of signs.) J(d)Finally, the gas is heated until its pressure increases from 1.5 atm to 6.0 atm at a constant volume. J(e)What is the net work done by the gas on its environment during the complete cycle described above? J
Question
A sample of gas, initially with a volume of 1.0 L, undergoes a thermodynamic cycle. Find the work done by the gas on its environment during each stage of the cycle described below. (Enter your answers in J.)(a)First, the gas expands from a volume of 1.0 L to 6.5 L at a constant pressure of 6.0 atm. J(b)Second, the gas is cooled at constant volume until the pressure falls to 1.5 atm. J(c)Third, the gas is compressed at a constant pressure of 1.5 atm from a volume of 6.5 L to 1.0 L. (Note: Be careful of signs.) J(d)Finally, the gas is heated until its pressure increases from 1.5 atm to 6.0 atm at a constant volume. J(e)What is the net work done by the gas on its environment during the complete cycle described above? J
Solution
(a) The work done by the gas during expansion can be calculated using the formula W = PΔV, where P is the pressure and ΔV is the change in volume. The pressure is given as 6.0 atm, but we need to convert this to pascals (Pa) because the standard unit of pressure in the SI system is Pa. 1 atm = 101325 Pa. So, 6.0 atm = 6.0 * 101325 = 607950 Pa. The change in volume ΔV = V_final - V_initial = 6.5 L - 1.0 L = 5.5 L. We also need to convert this to cubic meters (m^3), the SI unit of volume. 1 L = 0.001 m^3. So, 5.5 L = 5.5 * 0.001 = 0.0055 m^3. Substituting these values into the formula gives W = 607950 Pa * 0.0055 m^3 = 3343.725 J.
(b) The work done by the gas during cooling at constant volume is zero because there is no change in volume. W = PΔV = P * 0 = 0 J.
(c) The work done by the gas during compression can be calculated in the same way as in part (a). The pressure is now 1.5 atm = 1.5 * 101325 = 151987.5 Pa. The change in volume is ΔV = V_final - V_initial = 1.0 L - 6.5 L = -5.5 L = -0.0055 m^3. Substituting these values into the formula gives W = 151987.5 Pa * -0.0055 m^3 = -835.93125 J. The negative sign indicates that work is done on the gas.
(d) The work done by the gas during heating at constant volume is again zero because there is no change in volume. W = PΔV = P * 0 = 0 J.
(e) The net work done by the gas on its environment during the complete cycle is the sum of the work done during each stage: W_net = W_a + W_b + W_c + W_d = 3343.725 J + 0 J - 835.93125 J + 0 J = 2507.79375 J.
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