
When an ideal gas in a cylinder was compressed isothermally by a piston, the work done on the gas found to be \[1.5 \times {10^4}\;\]joules. During this process about
A) \[3.6 \times {10^{3\;}}cal\] of heat flowed out from the gas
B) \[3.6 \times {10^{3\;}}cal\] of heat flowed into the gas
C) \[1.5 \times {10^4}\;cal\] of heat flowed into the gas
D) \[1.5 \times {10^4}\;cal\] of heat flowed out from the gas
Answer
162.6k+ views
Hint:
An isothermal process is one in which the system's temperature stays constant. This indicates that the change in temperature is zero. So, heat is exchanged continuously between the system and surrounding to maintain the temperature of the system constant. So, the system's overall energy content is constant.
Complete step by step solution:
An isothermal process is a process where the temperature of the system remains constant.
According to the first law of thermodynamics, the change in internal energy is the function only of temperature.
Hence, \[\vartriangle U = f\left( T \right)\] where T is the temperature, \[\vartriangle U\] is change in internal energy.
If temperature is constant then \[\vartriangle U = 0\].
Now, \[\vartriangle Q = \vartriangle U + W\] here \[\vartriangle Q\]is the net heat transfer and W is network done.
\[ \Rightarrow \vartriangle Q = W\]
For an isothermal process, \[W = \smallint PdV\] here P is pressure and V is volume.
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 flowing out from the system.
Now calculate the amount of heat. We know that,
\[\Delta Q = \Delta U + \Delta W \Rightarrow \Delta W(\because \Delta U = 0)\]
\[ \Rightarrow \Delta Q = - 1.5 \times {10^4}J = 4.181.5 \times {10^4}cal\]
\[ = - 3.6 \times {10^3}cal\]
The negative sign shows that the heat is flowing out from the gas.
Hence A is the correct option.
Therefore, option (A) is the correct option.
Note:
When a gas is compressed isothermally then heat is rejected from the system. The first law of thermodynamics, the change in internal energy is the function only of temperature. The mathematical expression of laws of thermodynamics is , \[\vartriangle Q = \vartriangle U + W\] here \[\vartriangle Q\]is the net heat transfer and W is net work done , \[\vartriangle U\]= change in internal energy.
An isothermal process is one in which the system's temperature stays constant. This indicates that the change in temperature is zero. So, heat is exchanged continuously between the system and surrounding to maintain the temperature of the system constant. So, the system's overall energy content is constant.
Complete step by step solution:
An isothermal process is a process where the temperature of the system remains constant.
According to the first law of thermodynamics, the change in internal energy is the function only of temperature.
Hence, \[\vartriangle U = f\left( T \right)\] where T is the temperature, \[\vartriangle U\] is change in internal energy.
If temperature is constant then \[\vartriangle U = 0\].
Now, \[\vartriangle Q = \vartriangle U + W\] here \[\vartriangle Q\]is the net heat transfer and W is network done.
\[ \Rightarrow \vartriangle Q = W\]
For an isothermal process, \[W = \smallint PdV\] here P is pressure and V is volume.
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 flowing out from the system.
Now calculate the amount of heat. We know that,
\[\Delta Q = \Delta U + \Delta W \Rightarrow \Delta W(\because \Delta U = 0)\]
\[ \Rightarrow \Delta Q = - 1.5 \times {10^4}J = 4.181.5 \times {10^4}cal\]
\[ = - 3.6 \times {10^3}cal\]
The negative sign shows that the heat is flowing out from the gas.
Hence A is the correct option.
Therefore, option (A) is the correct option.
Note:
When a gas is compressed isothermally then heat is rejected from the system. The first law of thermodynamics, the change in internal energy is the function only of temperature. The mathematical expression of laws of thermodynamics is , \[\vartriangle Q = \vartriangle U + W\] here \[\vartriangle Q\]is the net heat transfer and W is net work done , \[\vartriangle U\]= change in internal energy.
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