
Two litres of ${{N}_{2}}$ at ${{0}^{0}}C$ and at 5 atm are expanded isothermally against a constant external pressure of 1 atm until pressure of gas reaches 1 atm. Assuring gas to be ideal calculates the work of expansion.
Answer
581.4k+ views
Hint: The answer to this question is based on the basic concept of physical chemistry that includes calculation using the formula ${{P}_{1}}{{V}_{1}} = {{P}_{2}}{{V}_{2}}$ that leads you to the correct answer by substituting values in actual formula $w = -{{P}_{ext}}({{V}_{2}}-{{V}_{1}})$
Complete Solution :
In our chapters of chemistry we have studied about various theories of gas and laws relating to this.
Here, the question is about the calculation of work of expansion and let us know what this means.
- We are familiar with the P – V diagram and the calculation of work based on this diagram.
Thus, here work of expansion is nothing but the area under the curve on a P – V diagram even if the process is not isobaric which describes the expansion.
- We have also studied about the Boyle’s law which is the formula given by ${{P}_{1}}{{V}_{1}}={{P}_{2}}{{V}_{2}}$ …….(1)
- Now according to the data given, ${{V}_{1}}=2L$, $T=273K$ and ${{P}_{1}} = 5atm$
- Now, after expansion, ${{V}_{2}}=?$, ${{P}_{1}} = 1atm$ and external pressure${{P}_{ext}} = 1atm$
They have given that the gas is expanding isothermally and the gas is ideal. Thus, the formula for ideal gas is given by, $PV=nRT$ where T and PV are constant here.
- Now we have to find ${{V}_{2}}$
From equation (1) ${{P}_{1}}{{V}_{1}} = {{P}_{2}}{{V}_{2}}$
Rearranging, ${{V}_{2}}=\dfrac{{{P}_{1}}{{V}_{1}}}{{{P}_{2}}} = \dfrac{5\times 2}{1}L$
Thus, ${{V}_{2}} = 10L$
- Now, work of expansion is given by, $w = -{{P}_{ext}}({{V}_{2}}-{{V}_{1}})$
Substituting the values, we get
$w = -1(10-2)L-atm = -8L-atm$
To convert this into joules, we have 1L-atm = 101.3 J
Therefore,
$w = -8\times 101.3J = -810.4J$
Thus, the correct answer is -810.4 J
Note: Changing the units from 1 L-atm to joules is that you have to multiple by the gas constant value in J/mol and then divide by the gas constant value in form of L-atm/mol K. These answers give the value 101.3J and remember this value which helps you for this unit conversion.
Complete Solution :
In our chapters of chemistry we have studied about various theories of gas and laws relating to this.
Here, the question is about the calculation of work of expansion and let us know what this means.
- We are familiar with the P – V diagram and the calculation of work based on this diagram.
Thus, here work of expansion is nothing but the area under the curve on a P – V diagram even if the process is not isobaric which describes the expansion.
- We have also studied about the Boyle’s law which is the formula given by ${{P}_{1}}{{V}_{1}}={{P}_{2}}{{V}_{2}}$ …….(1)
- Now according to the data given, ${{V}_{1}}=2L$, $T=273K$ and ${{P}_{1}} = 5atm$
- Now, after expansion, ${{V}_{2}}=?$, ${{P}_{1}} = 1atm$ and external pressure${{P}_{ext}} = 1atm$
They have given that the gas is expanding isothermally and the gas is ideal. Thus, the formula for ideal gas is given by, $PV=nRT$ where T and PV are constant here.
- Now we have to find ${{V}_{2}}$
From equation (1) ${{P}_{1}}{{V}_{1}} = {{P}_{2}}{{V}_{2}}$
Rearranging, ${{V}_{2}}=\dfrac{{{P}_{1}}{{V}_{1}}}{{{P}_{2}}} = \dfrac{5\times 2}{1}L$
Thus, ${{V}_{2}} = 10L$
- Now, work of expansion is given by, $w = -{{P}_{ext}}({{V}_{2}}-{{V}_{1}})$
Substituting the values, we get
$w = -1(10-2)L-atm = -8L-atm$
To convert this into joules, we have 1L-atm = 101.3 J
Therefore,
$w = -8\times 101.3J = -810.4J$
Thus, the correct answer is -810.4 J
Note: Changing the units from 1 L-atm to joules is that you have to multiple by the gas constant value in J/mol and then divide by the gas constant value in form of L-atm/mol K. These answers give the value 101.3J and remember this value which helps you for this unit conversion.
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