5 mol of an ideal gas is kept at 394◦C in an expansion from 1 L to 5 L .How much work is done by the gas? Answer in units of J.Given: R = 8.31 J/K/mol ,
Hmm. An isothermal expansion. The pressure must fall by a factor of 5 during this process. The initial pressure is
P1 = n R T = (5 moles/liter)*0.08206 Liter atm/mole K)* 667 K = 274 atm = 2.77*10^7 N/m^2
P2, the final pressure, is 1/5 of P1
The work done is the integral of P dV. Since PV is constant in an isothermal process, P = P1 V1/V
Work = Integral P1*V1 dV/V
V1=1 to V2=5
= 2.77*10^7 N/m^2 * 10^-3 m^3 * ln 5
= 4.46*10^4 Joules
Check my work. The method is more likely to be correct than my calculations
To find the work done by the gas, we can use the formula:
Work (W) = -PΔV
Where:
P = pressure of the gas
ΔV = change in volume
In this case, the gas is expanding from 1 L to 5 L. So, the change in volume (ΔV) = 5 L - 1 L = 4 L.
Now, let's find the pressure (P) of the gas using the ideal gas law:
PV = nRT
Where:
P = pressure
V = volume
n = number of moles
R = ideal gas constant (given as 8.31 J/K/mol)
T = temperature in Kelvin
Given:
n = 5 mol
R = 8.31 J/K/mol
T = 394°C = 394 + 273 = 667 K (convert to Kelvin)
Let's calculate the pressure (P):
P = (nRT) / V
P = (5 mol * 8.31 J/K/mol * 667 K) / 1 L
P = 27751 J/L
Now substitute the values into the work formula:
W = -PΔV
W = -27751 J/L * 4 L
W = -110,804 J
The work done by the gas is -110,804 J, where the negative sign indicates that work is done on the system (gas).