Advertisements
Advertisements
प्रश्न
For a given ideal gas 6 × 105 J heat energy is supplied and the volume of gas is increased from 4 m3 to 6 m3 at atmospheric pressure. Calculate
- the work done by the gas
- change in internal energy of the gas
- graph this process in PV and TV diagram
Advertisements
उत्तर
Heat energy supplied to gas Q = 6 × 105 J
Change in volume ∆V = (6 – 4) = 2 m3
1 atm = 1.013 × 105 Nm−2
a. Work done by the gas W = P × ∆V
= 1.013 × 105 × 2
= 2.026 × 105
W = 202.6 kJ
b. Change in internal energy of the gas
∆U = Q − P. ∆V
= 6 × 105 − (202.6 × 103)
= 6 × 105 − 2.026 × 105
= 3.974 × 105
∆U = 397.4 kJ
c.
APPEARS IN
संबंधित प्रश्न
Answer in brief.
Why should a Carnot cycle have two isothermal two adiabatic processes?
For work done to be reversible, the process should be ______
Draw a p-V diagram of the irreversible process.
3 mole of a gas at temperature 400 K expands isothermally from an initial volume of 4 litres to a final volume of 8 litres. Find the work done by the gas. (R = 8.31 J mol-1 K-1)
Explain in detail the isochoric process.
Draw the TP diagram (P-x axis, T-y axis), VT(T-x axis, V-y axis) diagram for
- Isochoric process
- Isothermal process
- Isobaric process
In an isothermal process, the volume of an ideal gas is halved. One can say that ____________.
We consider a thermodynamic system. If `Delta"U"` represents the increase in its internal energy and W the work done by the system, which of the following statements is true?
Explain the thermodynamic process.
When an inflated ballon is suddenly burst, why is the emerging air slightly cooled?
