Advertisements
Advertisements
प्रश्न
Give the expression for work done by the gas.
Advertisements
उत्तर
When a gas expands against pressure, it does work on the surroundings. The work done in expansion for volume V1 to V2 is given by
W = `int_("V"_1)^("V"_2) "PdV"`
If the pressure remains constant during expansion,
Then W = P(V2 – V1) = P∆V
If the volume remains constant, then W = 0.
If there is no external pressure, then no work is done. For example, when a gas expands freely in a vacuum, no work is done by it.
APPEARS IN
संबंधित प्रश्न
The change in internal energy in a reaction when 2kJ of heat is released by the system and 6 kJ of work is done on the system will be ______.
Define the internal energy of the system.
Can we measure the temperature of the object by touching it?
Give the sign convention for Q and W.
Define the quasi-static process.
Derive the expression for the work done in a volume change in a thermodynamic system.
An ideal gas is allowed to expand in a well insulated container against a constant external pressure of 2.5 bar from an initial volume of 2.50 dm3 to a final volume of 4.50 dm3. The change in internal energy (∆U) of the gas will be ______.
A gas is allowed to expand in a well insulated container against a constant external pressure of 2.5 Bar from an initial volume of 2.5L to a final volume of 4.5L The change in internal energy, ΔU of the gas will be ______.
The value of Δ U is ______ when 2 kJ of heat is released and 6 kJ of work is done on the system.
Calculate the change in internal energy of a gas if a gas is compressed adiabatically from 25 dm³ to 15 dm³ at constant external pressure of 4 bar.
