Question

3. n moles of an ideal gas undergo an isobaric process 1->2 and then the isochoric process 2->3 shown in Fig. 1 in such was that the gas performs work A. The ratio of P2 and P3 is known: P2/P3=k. The temperature T1 in the state 1 equals to the temperature T3 In state 3. Calculate temperature T3.

4. A monoatomic gas takes up a volume of V=4m3 and is at a pressure of 8x105 Pa. The gas undergoes an isothermal expansion reaching the final pressure of 1 atm. Calculate a) the work done to the gas in such a process b) the amount of heat absorbed by the gas c) change in the internal energy of the gas.

Answer 1

Question #81525, Physics / Molecular Physics | Thermodynamics

3. n moles of an ideal gas undergo an isobaric process 1->2 and then the isochoric process 2->3 shown in Fig. 1 in such was that the gas performs work A. The ratio of P2 and P3 is known: P2/P3=k. The temperature T1 in the state 1 equals to the temperature T3 In state 3. Calculate temperature T3.

Solution

1) $W = p_{2}(V_{2} - V_{1}) = A$

2) $nRT_{1} = p_{1}V_{1} = p_{2}V_{1}$

3) $\frac{p_3}{p_2} = \frac{T_3}{T_2} = \frac{1}{k} \rightarrow T_3 = \frac{T_2}{k}$

4) $nRT_{2} = p_{2}V_{2}$

Thus,

$$A = n R (T _ {2} - T _ {1})$$ $$T _ {2} = T _ {1} + \frac {A}{n R}.$$

So,

$$T _ {3} = \frac {1}{k} \Big (T _ {1} + \frac {A}{n R} \Big).$$

4. A monoatomic gas takes up a volume of $V = 4m3$ and is at a pressure of $8 \times 105 \, \text{Pa}$ . The gas undergoes an isothermal expansion reaching the final pressure of 1 atm. Calculate a) the work done to the gas in such a process b) the amount of heat absorbed by the gas c) change in the internal energy of the gas.

Solution

a) the work done to the gas:

$$W = n R T \ln \frac {V _ {2}}{V _ {1}}$$ $$T = \text{const} \rightarrow p _ {1} V _ {1} = p _ {2} V _ {2} \rightarrow \frac {V _ {2}}{V _ {1}} = \frac {p _ {1}}{p _ {2}}$$ $$n R T = p V = p _ {1} V _ {1}$$

Thus,

$$W = p _ {1} V _ {1} \ln \frac {p _ {1}}{p _ {2}} = (800000) (4) \ln \frac {8}{1.01325} = 6.6 \, \text{MJ}.$$

b) the amount of heat absorbed by the gas:

$$Q = W = 6.6 \, \text{MJ}.$$

c) change in the internal energy of the gas:

$$\Delta U = Q - W = 0 \, J.$$

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