Question

If an ideal gas compressed during isothermal process then:-
A) no work is done against gas
B) heat is rejected by gas
C) it's internal energy will increase
D) pressure does not change.

Answer 1

Hint:
The ideal gas law establishes a mathematical relationship between a gas's pressure, volume, and temperature. \[PV = nRT\] Here P is for pressure, V is for volume, n is for moles of gas, R is for the gas constant, and T is for temperature. This is how the ideal gas law is expressed. Temperature has only one function, which is internal energy.


Complete step by step solution:
A mathematical relationship between a gas's pressure, volume, and temperature is known as the ideal gas law. The ideal gas law is stated as \[PV = nRT\] , where P is the pressure, V is the volume, n is the number of moles of gas, R is the gas constant, and T is the temperature. When the situation is isothermal, which means that the temperature is constant then $PV = {\text{constant}}$.

In an isothermal process, the change in temperature is zero.

According to the first law of thermodynamics, the change in internal energy is the function only of temperature.
hence, \[\vartriangle U = f\left( T \right)\]
If temperature is constant then △U=0.
Now, \[\vartriangle Q = \vartriangle U + W\]
\[ \Rightarrow \vartriangle Q = W\]
For isothermal process, \[W = \smallint PdV\]
because compression is occurring then we can say that \[dV < 0\].
\[\therefore W < 0\] hence, \[\vartriangle Q < 0\;\]. Therefore we can say that heat is rejected from the system.
Option B is the correct


Therefore, option (B) is the correct option.



Note:
An isothermal process is one in which the system's temperature stays constant. A system's overall energy content is constant. Although energy can be moved between systems, the overall amount is constant. The change in internal energy is therefore solely dependent on temperature, as stated by the first law of thermodynamics.