1.6 mol of Neon gas are heated from 20.0 °C to 85.2 °C at a constant pressure of 1 atm. Calculate
the work done. (R = 0.082057 L.atm/ K. mol)
so the formula is W = nR∆T
Thank you for the answer above:)
W=-nRT
=-(1.6)(O.O82507)(65)
=-8.53J
Work done = (Pressure)*(Volume change)
= P(nR(T2-T1)]/P
= n*R(T2-T1)
= 1.6(0.082057 L.atm/ K. mol)*65.2K
= 8.56 L*atm
Those are inconvenient units for work, so multiply by 10^-3 m^2/L*(1.013*10^5 N/m^2/atm
I get 867 N-m (Joules)
To calculate the work done, we can use the formula:
Work = n * R * ΔT
where:
- n is the number of moles of the gas
- R is the ideal gas constant
- ΔT is the change in temperature in Kelvin
In this case, we are given:
- n = 1.6 mol
- R = 0.082057 L.atm/ K. mol
- ΔT = (85.2 °C - 20.0 °C) = 65.2 °C
However, we need to convert the temperature to Kelvin since the gas constant is given in terms of Kelvin. To convert from Celsius to Kelvin, we use the formula:
T(K) = T(°C) + 273.15
So, we have:
- T(initial) = 20.0 °C + 273.15 = 293.15 K
- T(final) = 85.2 °C + 273.15 = 358.35 K
- ΔT = T(final) - T(initial) = 358.35 K - 293.15 K = 65.2 K
Now, we can substitute the values into the formula:
Work = 1.6 mol * 0.082057 L.atm/K.mol * 65.2 K
Calculating the above expression gives the work done in terms of L.atm:
Work = 8.42 L.atm
So, the work done is 8.42 L.atm.