An ideal gas is sealed within a container at a tempreture of 17 degree celcies and a pressure of 101KPa. The container is heated until the tempreture of the gas reaches 100 degree celcies. A valve in the container is then opened to allow gas to escape until the pressure falls back to 101KPa at 100 degree celcies. 1) calculate the pressure in the gas just before the valve is opened. 2) calculate the fraction of the initial mass of gas that was lost as a result of opening the valve.

(a) p1/t1 = p2/t2

p2=p1(t2/t1)
p2=100kpa(373/290)
p2=101x10^3 pa (1.286)
p2=101x1286
p2=129,886pa or p2=129.886kpa
or in simple way
p2=101x10^3(1.286)
p2=101x10^3(1286x10^-3)
p2=101x1286
p2=p2=129,886pa or p2=129.886kpa
(b)

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To solve this problem, we will use the ideal gas law equation, which 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.

1) Calculate the pressure in the gas just before the valve is opened:

To solve for the initial pressure, we can use the ideal gas law equation and assume the initial volume and the number of moles remain constant:

P1 * V = n * R * T1 (equation 1)

Given:
T1 = 17 °C = 17 + 273 = 290 K
P1 = 101 kPa

We can rearrange equation 1 to solve for P1:

P1 = (n * R * T1) / V

Since the volume and the number of moles remain constant, we can assume that (n * R) / V is a constant, which we'll call C1. Thus, we can rewrite the equation as:

P1 = C1 * T1 (equation 2)

Now, substituting the values known:

P1 = C1 * T1
101 kPa = C1 * 290 K

Solving for C1:

C1 = 101 kPa / 290 K ≈ 0.348 kPa/K

Now, we can substitute T2 = 100 °C = 100 + 273 = 373 K into equation 2 to find the pressure just before the valve is opened:

P2 = C1 * T2
P2 = 0.348 kPa/K * 373 K
P2 ≈ 129.804 kPa

So, the pressure in the gas just before the valve is opened is approximately 129.804 kPa.

2) Calculate the fraction of the initial mass of gas that was lost as a result of opening the valve:

To calculate the fraction of the initial mass lost, we need to consider the relationship between pressure and the number of moles of gas, assuming the volume and temperature remain constant.

Using the ideal gas law equation, we know:

P * V = n * R * T

Since the volume and temperature remain constant:

P1 * V = n1 * R * T1 (equation 3)
P2 * V = n2 * R * T2 (equation 4)

Dividing equation 4 by equation 3 and rearranging, we get:

n2 / n1 = (P2 * T1) / (P1 * T2)

Substituting the given values:

n2 / n1 = (129.804 kPa * 290 K) / (101 kPa * 373 K)

Simplifying:

n2 / n1 ≈ 1.036

The fraction of the initial mass lost is approximately 1.036 - 1 = 0.036.

Therefore, approximately 0.036 or 3.6% of the initial mass of gas was lost as a result of opening the valve.