Question

a gas is compressed from a volume of 2m3 to a volume of 1m3 at a constant pressure of 100N/m2. Then it is heated at constant volume by supplying 150J of energy. As a result, the internal energy of the gas increases/decreases by??

Answer 1

Answer on Question #41632 – Physics – Molecular Physics | Thermodynamics

Question.

a gas is compressed from a volume of $2\mathrm{m}3$ to a volume of $1\mathrm{m}3$ at a constant pressure of $100\mathrm{N / m}2$ . Then it is heated at constant volume by supplying 150J of energy. As a result, the internal energy of the gas increases/decreases by??

$$V _ {1} = 2 m ^ {3}$$ $$V _ {2} = 1 m ^ {3}$$ $$P = 1 0 0 \frac {N}{m ^ {2}} = c o n s t$$ $$\delta Q _ {2} = 1 5 0 J$$ $$d U > 0 \text{ or } d U < 0?$$

Solution.

First law of thermodynamics:

$$d U = \delta Q - \delta A$$

$dU$ is an internal energy;

$\delta Q$ is a heat added to the gas;

$\delta A$ is a work done by the gas.

Consider two processes in this problem.

$$d U = d U _ {1} + d U _ {2}$$

In the first process (isobaric compression) the work was made on the gas:

$$\delta A = P d V = P \left(V _ {2} - V _ {1}\right) = - P \left(V _ {1} - V _ {2}\right) < 0 \rightarrow - \delta A = P \left(V _ {1} - V _ {2}\right) > 0$$ $$\delta Q = C _ {p} d T$$

Charles's law (for isobaric process):

$$\frac {V}{T} = c o n s t$$

Therefore, in our case $dV < 0 \rightarrow dT < 0 \rightarrow \delta Q = C_p dT < 0$

$$\mathrm {S o}, d U _ {1} = \delta Q - \delta A = C _ {p} d T + P (V _ {1} - V _ {2}) > 0$$

$C_p = \frac{i + 2}{2} R$, where $i$ is number of degrees of freedom. For example, for monoatomic gas $i = 3$ and for diatomic gas $i = 5$. For n-atomic gas $(n > 3)$ $i = 6$.

Ideal gas law: $PV = RT$. For isobaric process: $PdV = RdT$

So, $\delta Q = C_{p}dT = \frac{i + 2}{2} RdT = \frac{i + 2}{2} PdV = \frac{i + 2}{2} P(V_{2} - V_{1})$

$$d U _ {1} = \frac {i + 2}{2} P (V _ {2} - V _ {1}) - P (V _ {2} - V _ {1}) = \frac {i}{2} P (V _ {2} - V _ {1}) < 0$$

In the second process gas (isochoric heating) is heated at constant volume. Therefore,

$$d V = 0 \rightarrow \delta A = P d V = 0$$ $$d U _ {2} = \delta Q _ {2} > 0$$

So,

$$d U = d U _ {1} + d U _ {2} = \frac {i}{2} P (V _ {2} - V _ {1}) + \delta Q _ {2}$$

Calculate $dU$:

$$d U = \frac {i}{2} 100 (1 - 2) + 150 = - 50 \cdot i + 150 = - 50 (i - 3)$$

So, for monoatomic gas $(i = 3)$: $dU = 0$.

For diatomic gas $(i = 5)$: $dU = -100J < 0$

For n-atomic gas $(i = 6)$: $dU = -150J < 0$; $n > 3$

So, $dU$ doesn't increase. It remains constant for monoatomic gas and decreases for n-atomic gas.

Answer.

For monoatomic gas $(i = 3)$: $dU = 0$. It's remains constant

For diatomic gas $(i = 5)$: $dU = -100J < 0$. It decreases by $100J$

For n-atomic gas $(i = 6)$: $dU = -150J < 0$; $n > 3$. It decreases by $150J$

http://www.AssignmentExpert.com/