2 moles of ideal gas are in a cylinder container with cross-sectional area 1*10^-2 m^3. Cylinder is fitted with a light, movable piston that is open to atmosphere, where pressure is 10^-5 PA.A 50KG MASS PLACED ON PISTON KEEPS IT IN mechanical equilibrium when temperature is 300K. When temperature is reduced to 250K, find pressure.

Are you sure about that atmospheric pressure of 10^-5 Pa? It is much closer to 10^5 Pa.

It seems to me that the the pressure inside the piston is determined by the weight per area of the piston (and the 50 kg mass above it) and the atmospheric pressure above, and will not change.

To find the pressure when the temperature is reduced to 250K, we can use the ideal gas law, which states:

PV = nRT,

where:
P = pressure in Pascal (Pa),
V = volume in cubic meters (m³),
n = number of moles of gas,
R = ideal gas constant (8.314 J/(mol·K)),
T = temperature in Kelvin (K).

First, let's find the initial volume (V1) of the gas when the piston is at mechanical equilibrium. The cross-sectional area of the cylinder is given as 1 × 10^-2 m². Since the piston is open to the atmosphere, its weight is balanced by the atmospheric pressure of 10^-5 Pa. Therefore, the initial pressure (P1) is also 10^-5 Pa.

Using the given formula for pressure (P = F/A), we can calculate the force (F) exerted by the 50 kg mass on the piston:

F = m × g,

where:
m = mass in kilograms (kg),
g = acceleration due to gravity (9.8 m/s²).

F = 50 kg × 9.8 m/s² = 490 N.

Since pressure is force per unit area, we can now calculate the initial volume (V1):

P1 = F / A1,

where:
A1 = cross-sectional area of the cylinder.

V1 = A1 × h,

where:
h = initial height of the gas column in meters.

Combining these equations, we have:

P1 = F / (A1 × h) => h = F / (A1 × P1).

Substituting the given values:

h = 490 N / (1 × 10^-2 m² × 10^-5 Pa).

Now, we need to convert the given number of moles (n) to the actual number of gas molecules (N):

N = n × Avogadro's number,

where:
Avogadro's number = 6.022 × 10^23 molecules/mol.

So, N = 2 mol × 6.022 × 10^23 molecules/mol.

Next, let's find the final volume (V2) when the temperature is reduced to 250K. According to the ideal gas law, we have:

P2V2 = nRT2.

Since the number of moles (n) remains constant, we can rewrite this equation as:

V2 = (P1V1T2) / (P2T1).

Now, to find the final pressure (P2), we can rearrange the formula:

P2 = (P1V1T2) / (V2T1).

Substituting the given and calculated values:

T1 = 300K,
T2 = 250K,
P1 = 10^-5 Pa,
V1 = A1 × h,
V2 = (P1V1T2) / (P2T1).

Plug in the values and solve for P2:

250K × V2 × 10^-5 Pa / (1 × 10^-2 m² × h × 300K).

Simplifying the equation using the calculated values, we can find the final pressure (P2) when the temperature is reduced to 250K.