
Calculate work done during an isothermal reversible process when \[{\text{5mole}}\] ideal gas is expanded so that its volume is doubled at ${\text{400K}}$.
A.${\text{11}}{\text{.5Kj}}$
B.${\text{ - 344Kj}}$
C.$0$
D.${\text{ - 2}}{\text{.8Kj}}$
Answer
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Hint:The thermodynamic process in which the temperature of system remains constant is known as isothermal process, and when this process continues in infinitely slow motion with no change in its temperature and pressure it is known as isothermal reversible process.
Complete step by step answer:
Isothermal process: A thermodynamic process in which the overall temperature of the system remains constant is known as the isothermal process.
Isothermal reversible process: The process of expansion or contraction in which if the piston moves slowly such that at each instant the temperature and pressure of gas remains constant and obeys the ideal gas laws is known as isothermal reversible process.
The formula used to calculate work done in a thermodynamic process (isothermal reversible process) is as follows:
${\text{W = - 2}}{\text{.303nRT}} \times {\text{log}}\dfrac{{{V_2}}}{{{V_1}}}$
where, W represents the work done, ${{\text{V}}_{\text{1}}}{\text{ and }}{{\text{V}}_{\text{2}}}$ are the volumes.
n represents the number of moles, R is a gas constant and T is the temperature in Kelvin.
As we have given that volume is doubled; $ \Rightarrow \dfrac{{{V_2}}}{{{V_1}}} = \dfrac{2}{1}$
Now put all the provided values in the above formula to find out the work done.
${\text{W = - 2}}{\text{.303R}} \times {\text{5}} \times 400 \times {\text{log}}\dfrac{2}{1}$
${\text{W = - 19}} \times {\text{5}} \times 400 \times {\text{log2}}$ [We know that; $2.303 \times R = 19$ ]
${\text{W = - 19}} \times {\text{5}} \times 400 \times 0.3$ [And we also know that; $\log 2 = 0.3$ ]
$ {\text{W = - 11400 J}} \\
{\text{W = - 11}}{\text{.4 KJ}} \\ $
Therefore the work done during isothermal reversible process is $ - 11.4KJ$
Hence, option (A) is correct.
Note:
In chemistry sign convention matters a lot, if the energy leaves the system its sign is negative and if energy enters a system its sign is positive. Negative work indicates that the force component is opposite to displacement, in thermodynamics negative work means removal of energy from the system.
Complete step by step answer:
Isothermal process: A thermodynamic process in which the overall temperature of the system remains constant is known as the isothermal process.
Isothermal reversible process: The process of expansion or contraction in which if the piston moves slowly such that at each instant the temperature and pressure of gas remains constant and obeys the ideal gas laws is known as isothermal reversible process.
The formula used to calculate work done in a thermodynamic process (isothermal reversible process) is as follows:
${\text{W = - 2}}{\text{.303nRT}} \times {\text{log}}\dfrac{{{V_2}}}{{{V_1}}}$
where, W represents the work done, ${{\text{V}}_{\text{1}}}{\text{ and }}{{\text{V}}_{\text{2}}}$ are the volumes.
n represents the number of moles, R is a gas constant and T is the temperature in Kelvin.
As we have given that volume is doubled; $ \Rightarrow \dfrac{{{V_2}}}{{{V_1}}} = \dfrac{2}{1}$
Now put all the provided values in the above formula to find out the work done.
${\text{W = - 2}}{\text{.303R}} \times {\text{5}} \times 400 \times {\text{log}}\dfrac{2}{1}$
${\text{W = - 19}} \times {\text{5}} \times 400 \times {\text{log2}}$ [We know that; $2.303 \times R = 19$ ]
${\text{W = - 19}} \times {\text{5}} \times 400 \times 0.3$ [And we also know that; $\log 2 = 0.3$ ]
$ {\text{W = - 11400 J}} \\
{\text{W = - 11}}{\text{.4 KJ}} \\ $
Therefore the work done during isothermal reversible process is $ - 11.4KJ$
Hence, option (A) is correct.
Note:
In chemistry sign convention matters a lot, if the energy leaves the system its sign is negative and if energy enters a system its sign is positive. Negative work indicates that the force component is opposite to displacement, in thermodynamics negative work means removal of energy from the system.
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