ThermodynamicsMedium
Question
One mole of an ideal gas undergoes the following cyclic process.
(i) Isochoric heating from (P1, V1, T1) to double temperature.
(ii) Isobaric expansion to double volume. (iii) Linear expansion (on P–V curve) to (P1, 8V1).
(iv) Isobaric compression to initial state.
If T1 = 300 K, then the magnitude of net work done by the gas in the cyclic process is
Options
A.2400 cal
B.1200 cal
C.4800 cal
D.3600 cal
Solution
$w = - \left( P_{1}V_{1} + \frac{1}{2} \times P_{1} \times 6V_{1} \right) = - 4P_{1}V_{1}$
$= - 4nRT_{1} = - 4 \times 1 \times 2 \times 300\text{ cal} $$$= - 2400\text{ cal}$$
Create a free account to view solution
View Solution FreeMore Thermodynamics Questions
Standard entropy of X2, Y2 and XY2 are 60, 40 and 50 JK-1mol-1, respectively. For the reaction,, ᐃH = - 30 kJ, to ...A Carnot engine operates between 327°C and 117°C. If it absorbs 120 cal heat per cycle from the source, then the heat re...Three lawn chairs, one made up of aluminium (heat capacity = 0.90 J/K-g), one of iron (heat capacity = 0.45 J/K-g) and o...The entropy change in the fusion of one mole of a solid melting at 300 K (latent heat of fusion, 2930 J/mol) is...The enthalpy of vaporization of a liquid is 30 kJ mol-1 and entropy of vaporization is 75 J mol-1. The boiling point of ...