A gas with initial conditions p_i, V_i, and T_i expands isothermally until V_f=6 V_i. What are (a) T_f and (b) p_f?
1)T_f/T_i=
2)p_f/p_i=
I don't know how to solve this
A answered this a few days ago.
http://www.jiskha.com/display.cgi?id=1313210061
I saw that but I don't understand what the actual answer should be.
To solve this problem, we need to apply the ideal gas law and the conditions for an isothermal process.
The ideal gas law states: PV = nRT, where P is the pressure, V is the volume, n is the number of moles, R is the ideal gas constant, and T is the temperature.
For an isothermal process, the temperature remains constant (T_f = T_i).
Let's go step by step to solve the problem:
(a) To find T_f, we know that T_f = T_i (since it is an isothermal process). Therefore, T_f/T_i is equal to 1.
Therefore, T_f/T_i = 1.
(b) To find p_f, we need to use the ideal gas law:
p_i * V_i = p_f * V_f
Since V_f = 6 * V_i, we can substitute it into the equation:
p_i * V_i = p_f * (6 * V_i)
V_i cancels out on both sides:
p_i = 6 * p_f
Now, we can rearrange the equation to isolate p_f:
p_f = p_i / 6.
Therefore, p_f/p_i is equal to 1/6.
So, the answers to the questions are:
(a) T_f/T_i = 1
(b) p_f/p_i = 1/6.