ThermodynamicsHard
Question
A quantity of 70 calories of heat is required to raise the temperature of 2 mole of an ideal gas at constant pressure from 40°C to 50°C. The amount of heat required to raise the temperature of the same gas through the same range at constant volume is
Options
A.90 calorie
B.70 calorie
C.50 calorie
D.30 calorie
Solution
$C_{P,m} = \frac{q_{p}}{n.\Delta T} = \frac{70}{2 \times 10} = \frac{7}{2}\text{ cal}$
$\therefore C_{v,m} = \frac{7}{2} - 2 = \frac{3}{2}\text{ cal}$
Now, $q_{v} = n.C_{v,m}.\Delta T = 2 \times \frac{3}{2} \times 10 = 30\text{ cal}$
Create a free account to view solution
View Solution FreeMore Thermodynamics Questions
If a closed system has adiabatic boundaries, then at least one boundary must be...The value of log10K for a reaction A ⇋ B is ( Given : áƒfHo298K = - 54.07 kJ mol-1, áƒrSo298K = 10JK-1mol-1 and R = ...An ideal gas is expanded irreversibly from 5 L to 10 L against a constant external pressure of 1 bar. The value of heat ...One mole of a non-ideal gas undergoes a change of state (2.0 atm, 3.0 L, 95 K) with a change in internal energy, (ᐃ...An isolated system comprises of the liquid in equilibrium with vapours. At this stage, the molar entropy of the vapour i...