The work done by an ideal gas in an isothermal expansion from volume V1 to volum
ID: 1522718 • Letter: T
Question
The work done by an ideal gas in an isothermal expansion from volume V1 to volume V2 is given by the formula:
W=nRTln (V2/V1).
Standard atmospheric pressure (1atm) is 101.3 kPa. If 1.0L of He gas at room temperature (20°C) and 1.0 atm of pressure is compressed isothermally to a volume of 100 mL, how much work is done on the gas?
A) 5.6 kJ B) 4.7×10^2 J C) 4.7×10^2 kJ d) 2.3 × 10^2 kJ e) 2.3×10^2 J
.
2. A 4kg mass of metal of unknown specific heat at a temperature of 600 °J is dropped into 0.5 kg of ice and 0.5 kg of water both at 0°C. With no heat losses to the surrounding, the equilibrium temperature of the mixture is 85°C. Calculate the specific heat of the metal.
Explanation / Answer
(1)
p*v = nRT
101.3 * 10^3 * 1.0 * 10^-3 = nRT
nRT = 101.3
Work done by the gas,
W = n*R*T ln (V2/V1).
W = 101.3 * ln(100/1000)
W = 101.3 * ln(0.1)
W = - 2.3 * 10^2 J
Work is done on the gas = - Work done by the gas
Work is done on the gas = 2.3 * 10^2 J
(2)
Unit of temperature mentiond in the question is wrong !! i am assuming it to be 600 oC .
m = 4kg
c = ?
mi = 0.5 Kg = 500 g
mw = 0.5 Kg = 500g
Specific Heat of Water, Cw = 4.186 J/gm k
Specific Heat of ice, Ci = 2.108 J/gm k
Latent Heat of ice, Li = 333.55 J/gm
Tf = 85 °C
Heat Lost by metal = Heat Gained by ice & water
m*c* T = mi * Li + mi * Cw * T + mi * Cs * T
4000 * c * (600 - 85) = 500 * 333.55 + 500 * 2.108 * 85 + 500 * 4.186 * 85
c = 0.2108 J/gm k
Specific heat of metal, c = 0.2108 J/gm k or 210.8 J/kg k
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