A monoatomic ideal gas undergoes a process in which the ratio of p to V at any instant is constant and equals to 1. What is the molar heat capacity of the gas?
4R / 2
3R / 2
5R / 2s
Zero
For a reaction
∆H = 30 kJ mol -1 and ∆S = 0.07 kJ K -1 mol -1 at 1 atm. The temperature up to which the reaction would not be spontaneous is
T < 400.08 K
T < 273.15 K
T < 428.57 K
T < 473.50 K
The molar heat capacity of water at constant pressure is 75 JK -1 mol -1. When 1.0 kJ of heat is supplied to 100 g of water, which is free to expand, the increase in temperature of water is
1.2 K
2.4 K
4.8 K
6.8 K
Mercury thermometer can be used to measure temperature up to
260oC
100oC
360oC
500oC
When the heat of a reaction at constant pressure is -2.5 x 103 cals and entropy change for the reaction is
7.4 cal deg -1, it is predicted that the reaction at 25oC is
Reversible
Spontaneous
Non-spontaneous
Irreversible
Statement I All exothermic reactions are spontaneous at room temperature.
Statement II In (∆G = ∆H - T∆S), ∆G becomes negative and negative sign of ∆G indicates spontaneous reaction.
Statement I is true, statement II is true, statement II is a correct explanation for statement I
Statement I is true, Statement II is true, statement II is not a correct explanation for statement I
Statement I is true; Statement II is false
Statement I is false; statement II is true.
In conversion of limestone to lime,
CaCO3 (s) → CaO(s) + CO2 (g)
The value of ∆Ho and ∆So are +179.1 kJ mol -1 and 160.2 J/ K respectively at 298 K and 1 bar. Assuming that ∆Ho and ∆So do not change with temperature, temperature above which conversion of limestone to lime will be spontaneous is
1008 K
1200 K
845 K
1118 K
Standard entropy of X2, Y2 and XY3 are 60, 40 and 50 JK -1 mol -1 respectively. For the reaction to be at the equilibrium, the temperature will be
750 K
1000 K
1250 K
500 K
2 moles of an ideal gas at 27oC are expanded reversibly from 2 L to 20 L. Find entropy change. (R = 2 cal/ mol K)
0
4
9.2
92.0
The value of log10 K for a reaction A B is
(Given, ∆rHo298 K = - 54.07 kJ mol -1
∆rSo298 k = 10 JK -1 mol -1
and R = 8.314 JK -1 mol -1
= 2.303 x 8.314 x 298
= 5705
5
10
95
100