Question

On an adiabatic process of an ideal gas pressure, volume, and temperature change such that pVγ is constant with γ=5/3 for a monatomic gas such as helium and γ=7/5 for a diatomic gas such as hydrogen at room temperature. Use numerical values to plot two isotherms of 1 mol of helium gas using the ideal gas law and two adiabatic processes mediating between them. Use T1=500K, V1=1L, and T2=300K for your plot.

(a) Plot A: Isotherm at 500 K, Plot B: Adiabatic Process 1, Plot C: Adiabatic Process 2, Plot D: Isotherm at 300 K
(b) Plot A: Isotherm at 300 K, Plot B: Adiabatic Process 1, Plot C: Adiabatic Process 2, Plot D: Isotherm at 500 K
(c) Plot A: Isotherm at 500 K, Plot B: Isotherm at 300 K, Plot C: Adiabatic Process 1, Plot D: Adiabatic Process 2
(d) Plot A: Adiabatic Process 1, Plot B: Isotherm at 500 K, Plot C: Isotherm at 300 K, Plot D: Adiabatic Process 2

Answer 1

To plot the isotherms and adiabatic processes for 1 mol of helium gas, we can use the ideal gas law and the given values. The equations PV = nRT and PV^γ = constant can be used to determine the corresponding volumes at different pressures.

Step-by-step explanation:

To plot the isotherms and adiabatic processes for 1 mol of helium gas, we can use the ideal gas law and the given values. Let's start with Plot A: Isotherm at 500 K. Using the ideal gas law, we have PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the gas constant, and T is the temperature. Rearranging the equation, we get V = (nRT)/P. Plugging in the values, we have V = (1 mol * 8.31 J/mol K * 500 K)/(P). Since we are looking for the isotherm at 500 K, we can use this equation to determine the corresponding volumes at different pressures.

For Plot B: Adiabatic Process 1, we know that PV^γ is constant, where γ is the ratio of specific heat capacities for the gas. For helium, γ = 5/3. We can use this equation to find the corresponding volumes at different pressures.

Similarly, for Plot C: Adiabatic Process 2, we can use the same equation with γ = 5/3. Finally, for Plot D: Isotherm at 300 K, we can use the ideal gas law with T = 300 K to determine the corresponding volumes at different pressures.