Question

1. An ideal monatomic gas, with 24.05 moles, expands adiabatically from 0.500m^3 to 1.75 m^3. IF the initial pressure and temperature are 1.40x10^5 Pa and 350K, respectively, find the change in internal energy of the gas if the final temperature of the gas is 152K.

2. A fridge does 17.5 KJ of work while moving 120KJ of thermal energy from inside the fridge. Calculate the fridge's coefficient of performance.

Answer #1

(1) Ratio of the two specific heats for an ideal monoatomic gas = = 1.67.

Initial pressure = P_{i} = 1.4 x 10^{5} Pa, and,
initial volume = V_{i} = 0.5 m^{3}.

Hence, P_{i}V_{i}^{}
= 1.4 x 10^{5} x 0.5^{1.67} = 4.4 x 10^{4}
= K, say.

Final volume of the gas = V_{f} = 1.75
m^{3}.

Hence, adiabatic work done = W = K ( V_{f}^{1 -
} - V_{i}^{1 -
} ) / ( 1 -
)

or, W = 4.4 x 10^{4} [ ( 1.75^{1 - 1.67} -
0.5^{1 - 1.67} ) / ( 1 - 1.67 ) ] J

or, W = 59350 J.

there is no heat exchange in an adiabatic process, so, heat input = Q = 0 J.

Hence, from the first law of thermodynamics,

change in the internal energy = | Q - W | = **59350
J**.

(2) Coeffitient of performance of the fridge = Heat taken out /
Work done = 120 kJ / 17.5 kJ = **6.875**.

3. An ideal monatomic gas expands isothermally from .500 m3 to
1.25 m3 at a constant temperature of 675 K. If the initial pressure
is 1.00 ∙ 105 Pa, find (a) the work done by the gas, (b) the
thermal energy transfer Q, and (c) the change in the internal
energy.

An ideal monatomic gas expands isothermally from 0.600 m3 to
1.25 m3 at a constant temperature of 730 K. If the initial pressure
is 1.02 ? 105 Pa find the following.
(a) the work done on the gas
J
(b) the thermal energy transfer Q
J
(c) the change in the internal energy
J

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K
(b) Compare your result to the result you would get if the gas
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Specify whether the change is an increase or a decrease.

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Considering 115 moles of gas with an initial pressure of 350 kPa
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