An ideal gas is sealed within a container at a temperature of 17°C and a pressure of 101 kPa. The container is heated until the temperature of the gas reaches100°C. A valve in the container is then opened to allow gas to escape until the pressure falls back to 101 kPa at 100°C. 

a) Calculate the pressure in the gas just before the valve is opened. 
b) Calculate the fraction of the initial mass of gas that was lost as a result of opening the valve

a. P2=T2P1/T1 temps in Kelvins.

b. n1=P1V1/RT1
n2=P2V2/RT2

n2/n2=T1/T2 in kelvins
that is the fraction that remains, so fraction lost is 1- that.

An ideal gas is sealed in a rigid container at initial temperature of 17 degree and pressure 100kap when it is heated it's temperature reaches 127 degree, the valve is operated. To allow. Gas to escape until the pressure falls back to 100kap at 127 degree. Calculate

are correct but what about the rest question.?

To solve this problem, we can use the ideal gas law equation: 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 in Kelvin.

a) To calculate the pressure in the gas just before the valve is opened, we need to convert the temperatures to Kelvin and use the ideal gas law equation.

Given:
Initial temperature (T1) = 17°C = 17 + 273.15 = 290.15 K
Initial pressure (P1) = 101 kPa
Final temperature (T2) = 100°C = 100 + 273.15 = 373.15 K
Final pressure (P2) = 101 kPa

Since we know the initial and final temperatures and pressures, we can set up the equation as follows:

P1V1 / T1 = P2V2 / T2

Since the volume remains constant, we can eliminate V1 and V2 from the equation.

P1 / T1 = P2 / T2

Now, substitute the values:

P1 = 101 kPa
T1 = 290.15 K
T2 = 373.15 K

101 kPa / 290.15 K = P2 / 373.15 K

Solving for P2:

P2 = (101 kPa / 290.15 K) * 373.15 K
P2 ≈ 129.6 kPa

Therefore, the pressure in the gas just before the valve is opened is approximately 129.6 kPa.

b) To calculate the fraction of the initial mass of gas that was lost as a result of opening the valve, we need to assume that the system is in equilibrium before and after the gas escapes.

Using Dalton's Law of Partial Pressures, we can write:

P_initial = P_gas + P_water vapor

Since the pressure of water vapor remains constant during this process, the initial pressure of the gas is equal to its partial pressure. Therefore:

P_initial = P_gas

After the gas escapes, the final pressure is still 101 kPa. So, the fraction of the initial mass lost can be calculated by the equation:

Fraction of initial mass lost = (P_initial - P_final) / P_initial

We have already determined that P_initial ≈ 129.6 kPa and P_final = 101 kPa.

Substituting in the values:

Fraction of initial mass lost = (129.6 kPa - 101 kPa) / 129.6 kPa

Fraction of initial mass lost ≈ 0.22 or 22%

Therefore, approximately 22% of the initial mass of gas was lost as a result of opening the valve.