Calculates the work done (in joules) when 42.8 g of tin dissolves in excess acid at 1.10 atm and 23 degrees Celcius. Assume ideal gas behaviour.

See your post below.

To calculate the work done when tin dissolves in excess acid, we need to determine the change in volume of the system.

We can use the ideal gas law equation to calculate the change in volume:

PV = nRT

Where:
P = pressure (in atmospheres)
V = volume (in liters)
n = number of moles
R = gas constant (0.0821 L·atm/(mol·K))
T = temperature (in Kelvin)

First, let's find the number of moles of tin (Sn) using its molar mass. The molar mass of tin is 118.71 g/mol.

molar mass of Sn = 118.71 g/mol
mass of Sn = 42.8 g

n = mass of Sn / molar mass of Sn
n = 42.8 g / 118.71 g/mol

Next, we need to convert the given temperature from Celsius to Kelvin:

T(°C) = T(K) - 273.15
T(°C) = 23 °C
T(K) = 23 + 273.15 K

Now, we can substitute the given values into the ideal gas law equation:

PV = nRT

(Volume of the system is not given, so we'll assume it is constant.)

P = 1.10 atm
V = unknown
n = (42.8 g / 118.71 g/mol)
R = 0.0821 L·atm/(mol·K)
T = (23 + 273.15) K

We can solve the equation for V:

V = (nRT) / P

Now, plug in the values:

V = [(42.8 g / 118.71 g/mol) * (0.0821 L·atm/(mol·K)) * (23 + 273.15) K] / 1.10 atm

Simplify the expression:

V = (42.8 * 0.0821 * 296.15) / 1.10

Now, we can calculate the volume (V) in liters.

V = 10.1167 L

The work done (W) can be determined using the equation:

W = -PΔV

Where:
W = work done (in joules)
P = pressure (in atmospheres)
ΔV = change in volume (in liters)

Let's determine the change in volume (ΔV):

ΔV = Vfinal - Vinitial
ΔV = V - Vinitial

Since the volume of the system was assumed to be constant, there is no change in volume, so ΔV = 0.

Now, substituting the values into the work equation:

W = -PΔV
W = -(1.10 atm) * (0 L - 10.1167 L)

W = -1.10 atm * -10.1167 L

W ≈ 11.13 J

Therefore, the work done when 42.8 g of tin dissolves in excess acid at 1.10 atm and 23 degrees Celsius is approximately 11.13 Joules.