Question

**One mole of nitrogen gas confined within a cylinder by a
piston is heated from 0**°**C to
852**°**C at** 1.00 atm.

**(a) Calculate the work of expansion of the gas in
joules** (1 J = 9.87 × 10^{−3} L·atm).
**Assume all the energy is used to do work. (Enter your
answer in scientific notation.)**

**(b) What would be the temperature change if the gas were
heated with the same amount of energy in a container of fixed
volume? (Assume the specific heat capacity of
N**_{2} **is 1.00
J/g**·**K.)**

Answer #1

SOLUTION:

(A) Work (W) = PΔV

P = pressure, ΔV = change in volume = V2 - V1

V1 = 22.4L because 1mol of gas at STP has volume of 22.4L. Gas
is initialy at 0^{o}C and 1atm i.e, at STP

P_{1}V_{1} / T_{1} =
V_{2}P_{2}/T_{2}

V_{2} = V_{1} X T_{2} / T_{1}
Because P1 = P2 = 1atm, T1 = 0C = 273K , T2 = 852C = 1125K

V_{2} = 22.4 L X 1125K / 273K = 92.30L

W = PΔV = 1atm X (92.3 - 22.4) = 69.9Latm = 69.9Latm / 9.87 ×
10^{−3} = 7.08 X 10^{3} J

(B) If the same energy is supplied to 1 mol of N2 the temperature will rise and can be calculated as:

Q = mcΔT

Q = heat supplied , m = mass of gas = 28g, c = specific heat =
**1.00 J/g**·**K**, ΔT = rise in
temperature = T2 - T1

T1 = 0^{o}C = 273 K, T2 = ?

7.08 X 10^{3} J = 28 X 1 X (T2 - 273)

(T2 - 273) = 7.08 X 10^{3} J / 28 = 252.85

T2 = 252.85 + 273 = 525.85K

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