Looking for classes? Ksquare Career Institute, Bengaluru →
During an isothermal expansion, a confined ideal gas does −150 J of work against its surroundings. This implies that
A
150 J of heat has been added to the gas
B
150 J of heat has been removed from the gas
C
300 J of heat has been added to the gas
D
No heat is transferred because the process is isothermal
Detailed Solution
First law of thermodynamics: $Q = \Delta U + W$, where W is the work done by the gas.
For an ideal gas the internal energy depends only on temperature; in an isothermal process $\Delta T = 0$, so $\Delta U = 0$.
Given W = −150 J.
$Q = 0 + (-150) = -150$ J
The negative sign means heat has left the gas.
Hence 150 J of heat has been removed from the gas.
For an ideal gas the internal energy depends only on temperature; in an isothermal process $\Delta T = 0$, so $\Delta U = 0$.
Given W = −150 J.
$Q = 0 + (-150) = -150$ J
The negative sign means heat has left the gas.
Hence 150 J of heat has been removed from the gas.