A gas is confined to a vertical cylinder by a piston of mass 2 kg and radius 1 cm. When 5J of heat are added, the piston rises by 2.4 cm. Find: (a) the work done by the gas; (b) the change in its internal energy. Atmospheric pressure is 105Pa.
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The work done by the gas is 0.79128 j
and the change in its internal energy is 4.208 j
Explanation:
note 1 : the question says atmospheric pressure to be 105Ka , it should be 105 kPa
as the work is done at the constant external pressure 105 Kpa
work done = P external * (Δ volume ) ..................... equation 1
Δ volume = π r ² Δ h ..... ..... eq 2
Δ h is the change is height in our case piston rises by 2.4 cm so Δ h is +2.4cm
Δ volume= 3.14 * (0.01)²* 0.024 //(be careful to change units in meter )
Δ volume = 0.000007536 m³
using above Δ volume in equation 1
work done BY THE GAS = 105000*0.000007536
= 0.79128 j
Using first law of thermodynamics :
Δ U = Δ Q + W
= 5 - 0.79128
// AS THE work is done by the gas we will use negative 0.79128
Δ U = 4.208 j
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