Solving Pressure and Force in Gas Experiments

[503gurl]

Homework Statement

a) Suppose that in the experiment involving P vs. T, the slope of the graph of P vs. T is 235 Pa/L/ Determine the pressure of the gas at T = 0 degrees celsius.

b) The radius of the syringe is approximately 5 mm. Suppose that the pressure of the gas inside the syringe is 2.00 x 10^5 Pa. Determine the force exerted by the gas on the plunger and the force exerted by your hand on the plunger to hold it in equilibrium.

c) Two groups perform the part of the lab involving P vs. V but start with different volumes of air in the syringe. The temperatures of the gases are the same for both groups. Group A attached the pressure sensor when the volume of the syringe is 20 ml and group B when the volume is 10 ml. Using Eq. (1) determine an expression for the slope of these graphs. Identify which quantities in this expression are different for the two groups and use this to relate the slopes of the graphs of P vs. 1/V for the two groups.

Homework Equations

PV= nRT

The Attempt at a Solution

 

[dikmikkel]

You know the ideal gas law how about dividing it by V then you have the slope. Also the important part of this question is to calculate in K 0 deg cel. is 273.14K

 

[503gurl]

So the equation I would use is PV = nRT/V ? Is this the right direction?

 

[I like Serena]

Try P=(nR/V) T.

The slope is (nR/V).

 

[asteriatic]

a) Using the ideal gas law, PV = nRT, we can rearrange the equation to solve for pressure (P). At T = 0 degrees Celsius, the temperature (T) is equal to 273 Kelvin. We can assume that the volume (V) remains constant, so we can set it equal to 1 liter (L). Therefore, we have:

P = (nRT)/V

Substituting the given slope of 235 Pa/L for P and the known values of n (number of moles), R (ideal gas constant), T (temperature), and V (volume), we can solve for the pressure:

235 Pa/L = (n x 8.31 J/mol*K x 273 K)/1 L

n = 0.0107 moles

Now, we can plug this value of n into the original equation to solve for P:

P = (0.0107 moles x 8.31 J/mol*K x 273 K)/1 L

P = 23.4 Pa

Therefore, the pressure of the gas at T = 0 degrees Celsius is 23.4 Pa.

b) To determine the force exerted by the gas on the plunger, we can use the equation F = PA, where P is the pressure and A is the area. The radius of the syringe is given to be 5 mm, which is equal to 0.005 m. We can use this to calculate the area of the plunger:

A = πr^2 = 3.14 x (0.005 m)^2 = 7.85 x 10^-5 m^2

Plugging in the given pressure of 2.00 x 10^5 Pa, we have:

F = (2.00 x 10^5 Pa)(7.85 x 10^-5 m^2) = 15.7 N

Therefore, the force exerted by the gas on the plunger is 15.7 N.

To calculate the force exerted by the hand on the plunger, we can use Newton's second law, Fnet = ma. Since the plunger is in equilibrium, the net force on it is 0 N. Therefore, the force exerted by the hand must be equal and opposite to the force exerted by the gas:

Fhand = -Fgas = -15.7 N

c) The ideal