One mole of an ideal gas does 3000 J of work on its surroundings as it expands isothermally to a final pressure of 1.00 atm and volume of 250 L. What is the initial volume and the temperature of the gas?

Question

Julia Aguilar · Accepted Answer

First, let's note a few important conditions:

Isothermal means the temperature is held constant, so #color(green)(DeltaT = 0)#.
In these conditions, there is no change in enthalpy #DeltaH# nor any change in internal energy #DeltaU#. Not sure if we need it, but we should know that #color(blue)(DeltaH = DeltaU = 0)# for an ideal gas if #DeltaT = 0#.
- We are given final pressure but not initial pressure, and final volume but not initial volume. So, we would probably have to manipulate the Ideal Gas Law #PV = nRT# for #T#, and maybe something to do with the work function for #V_1#.
  REVERSIBLE WORK VS VOLUME
  
  Now, we should know that reversible (efficient) work is defined as:
  
  #\mathbf(w_"rev" = -intPdV)#
  
  For this problem, the depiction in a PV-diagram goes like this:
  
  However, since we do not know #DeltaP# and the equation had assumed that #DeltaP = 0# (note that we are integrating with respect to volume, not pressure), we have use the Ideal Gas Law to rewrite the above equation as:
  
  #w_"rev" = -int_(V_1)^(V_2)(nRT)/VdV#
  
  Since #DeltaT = 0#, #nRT# is a constant. Let's pull it out of the integral.
  
  #w_"rev" = -nRTint_(V_1)^(V_2)1/VdV#
  #= -nRTln|V_2/V_1|#
  
  At this point, the shape of the PV-diagram curve looks sensible; it resembles the shape of a #-lnx# curve.
  
  THE INITIAL VOLUME FOR THE EXPANSION
  
  We are given the work, so we should be able to solve for the initial volume, #V_1#, using this equation. Since expansion work has been done by the gas, meaning that #V_2 > V_1#, we know that #color(green)(w_"rev" < 0)#, numerically.
  
  That makes sense because #ln|V_2/V_1|# when #V_2/V_1 > 1# is positive. Now let's get an expression for #V_1#.
  
  #-w_"rev"/(nRT) = ln|V_2/V_1|#
  #e^(-w_"rev""/"nRT) = V_2/V_1#
  #e^(w_"rev""/"nRT) = V_1/V_2#
  #color(green)(V_1 = V_2e^(w_"rev""/"nRT))#
  
  However, we do not know the temperature yet, so we'll have to put off calculating the initial volume for a bit longer.
  
  THE TEMPERATURE FOR THE EXPANSION
  
  Something we do know is that the temperature remained constant, so #T_2 = T_1#. Thus, we have these two relationships that cover the initial and final states:
  
  #P_1V_1 = color(blue)(nR)T#
  #color(blue)(P_2V_2) = color(blue)(nR)T#
  
  We know the values of what is in blue, which is enough.
  
  #color(blue)(T) = (P_2V_2)/(nR) = (("1 atm")("250 L"))/(("1 mol")("0.082057 L"cdot"atm/mol"cdot"K"))#
  #=# #color(blue)("3046.66 K")#
  
  Now we can find the initial volume:
  
  #color(blue)(V_1) = ("250 L")e^((-"3000 J")"/"[("1 mol")("8.314472 J/mol"cdot"K")("3046.66 K")]#
  #=# #color(blue)("222.08 L")#
  
  (As an aside, if you were curious, the initial pressure was about #"1.126 atm"#, from the Ideal Gas Law.)

Adrian Ayala · Answer

undefined The initial volume ($ V_i $) and the temperature ($ T $) of the gas can be calculated using the ideal gas law and the work done equation for an isothermal process.

Given:
- Work done ($ W $) = 3000 J
- Final pressure ($ P_f $) = 1.00 atm
- Final volume ($ V_f $) = 250 L

Using the ideal gas law $ PV = nRT $, where $ n $ is the number of moles, we can rearrange it to find $ n $:
$$ n = \frac{PV}{RT} $$

Since the process is isothermal, the temperature ($ T $) remains constant. Thus, $ T $ is the same at both initial and final states.

Then, we can find the initial volume ($ V_i $) using the work done equation for an isothermal process:
$$ W = nRT\ln\left(\frac{V_f}{V_i}\right) $$

Rearranging this equation to solve for $ V_i $, we get:
$$ V_i = V_f \exp\left(-\frac{W}{nRT}\right) $$

Substituting the values and solving for $ V_i $:
$$ V_i = 250 \times \exp\left(-\frac{3000}{nRT}\right) $$

Once $ V_i $ is found, you can use the ideal gas law to find the temperature $ T $ at either the initial or final state.

HIX Tutor · Answer

undefined To find the initial volume and the temperature of the gas, follow these steps:

### Step 1: Use the work equation for isothermal expansion.
The work done by an ideal gas during isothermal expansion is given by:
$$ 
W = -P_{\text{external}} \Delta V 
$$
Where:
- $ W $ is the work done (3000 J)
- $ P_{\text{external}} $ is the external pressure (1.00 atm)
- $ \Delta V $ is the change in volume

### Step 2: Convert pressure to the correct units.
Convert 1.00 atm to Joules per liter:
$$ 
1 \text{ atm} = 101.325 \text{ J/L} 
$$
So,
$$ 
P = 101.325 \text{ J/L} 
$$

### Step 3: Calculate the change in volume.
Using $ W = -P \Delta V $:
$$
-3000 \text{ J} = -101.325 \text{ J/L} \Delta V 
$$
Now solve for $ \Delta V $:
$$
\Delta V = \frac{3000 \text{ J}}{101.325 \text{ J/L}} \approx 29.6 \text{ L} 
$$

### Step 4: Find the initial volume.
We know the final volume is 250 L:
$$
V_f = 250 \text{ L} 
$$
Now, use the change in volume:
$$
\Delta V = V_f - V_i 
$$
So,
$$
29.6 \text{ L} = 250 \text{ L} - V_i 
$$
Rearranging gives:
$$
V_i = 250 \text{ L} - 29.6 \text{ L} \approx 220.4 \text{ L} 
$$

### Step 5: Find the temperature using the ideal gas law.
Use the ideal gas law:
$$
PV = nRT 
$$
We need to find $ T $. Rearranging gives:
$$
T = \frac{PV}{nR} 
$$
Where:
- $ n = 1 \, \text{mol} $
- $ R = 0.0821 \, \text{L atm/(K mol)}$

### Step 6: Calculate the temperature.
Now plug in the values:
Using final volume $ V_f = 250 \text{ L} $:
$$
T = \frac{(1 \, \text{atm} )(250 \text{ L})}{(1 \, \text{mol})(0.0821 \, \text{L atm/(K mol)})}
$$
Calculating gives:
$$
T \approx \frac{250}{0.0821} \approx 3041.7 \text{ K} 
$$

### Summary:
- **Initial Volume**: $ \approx 220.4 \text{ L} $
- **Temperature**: $ \approx 3041.7 \text{ K} $

If you have any questions, feel free to ask!