An ideal gas is in equilibrium at initial state with temperature T=137oC, pressure P = 0.75Pa and volume V = 0.75 m3. If there is a change in state in which the gas undergoes an isothermal process to a final state of equilibrium during which the volume is doubled. Calculate the temperature and pressure of the gas at this final state.
1.00pa
To solve this problem, we can apply the ideal gas law and use the fact that the process is isothermal, meaning the temperature remains constant.
The ideal gas law states:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant
T = temperature
First, let's calculate the number of moles of the gas in the initial state using the ideal gas law:
n = PV / RT
Given:
P = 0.75 Pa
V = 0.75 m^3
T = 137°C = 410 K (convert Celsius to Kelvin)
Let's calculate n:
n = (0.75 Pa) * (0.75 m^3) / (R * 410 K)
Now, since we have the number of moles, we can calculate the final pressure using the ideal gas law again:
P_final = (n * R * T) / V_final
Given:
V_final = 2 * V
Let's calculate P_final:
P_final = (n * R * T) / (2 * V)
Now, we need to calculate the final temperature. Since the process is isothermal, the temperature remains constant. Therefore, T_final = T.
Finally, substitute the values into the equations to get the final temperature and pressure:
T_final = T = 410 K
P_final = (n * R * T) / (2 * V)
Substitute the values of n, R, T, and V into the equation, and calculate P_final.