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Hrishant Goswami Grade: 10
        A sample of air weighing 1.18 g occupies 1.0 × 103 cm3 when kept at 300 K and 1.0 × 105 Pa. When 2.0 cal of heat of added to it at constant volume, its temperature increases by 1°C. Calculate the amount of heat needed to increase the temperature of air by 1°C at constant to increase if the temperature of air by 1C at constant pressure if the mechanical equivalent of heat is 4.2 × 107 erg/cal. Assume that air behaves as an ideal gas.
3 years ago

Answers : (4)

Aditi Chauhan
askIITians Faculty
396 Points
										Sol. m = 1.18 g, V = 1 × 10^3 cm^3 = 1 L T = 300 k, P = 10^5 Pa

PV = nRT or n = PV/RT = 10^5 = atm.
N = PV/RT = 1/8.2 * 10^-2 * 3 * 10^2 = 1/8.2 * 3 = 1/24.6
Now, C base v = 1/n = 24.6 * 2 = 49.2
C base p = R + C base v = 1.987 + 49.2 = 51.187
Q = nC base pdT = 1/24.6 * 51.187 * 1 = 2.08 Cal.
3 years ago
Deepak Patra
askIITians Faculty
474 Points
										Sol. C base P/C base V = 7.6, n = 1 mole, ∆T = 50K

(a) Keeping the pressure constant, dQ = du + dw,
∆T = 50 K, γ = 7/6, m = 1 mole,
dQ = du + dw ⇒ nC base VdT = du + RdT ⇒ du = nCpdT – RdT
= 1 * Rγ/γ – 1 * dT – RdT = R *7/6/7/6-1 dT – RdT
= DT – RdT = 7RdT = 6 RdT = 6 * 8.3 * 50 = 2490 J.
(b) Kipping Volume constant, dv = nC base VdT
= 1 * R/γ – 1 * dT = 1 * 8.3/7/6-1 * 50
= 8.3 * 50 * 6 = 2490 J
(c)Adiabetically dQ = 0,
= [n * R/γ – 1(T base 1 – T base 2)] = 1 * 8.3/7/6 -1 (T base 1 – T base 2)] = 8.3 * 50 * 6 = 2490 J
3 years ago
Jitender Pal
askIITians Faculty
365 Points
										Sol. . m = 1.18 g, V = 1 × 10^3 cm^3 = 1 L T = 300 k, P = 10^5 Pa

PV = nRT or n = PV/RT = 10^5 = atm.
N = PV/RT = 1/8.2 * 10^-2 * 3 * 10^2 = 1/8.2 * 3 = 1/24.6
Now, C base v = 1/n = 24.6 * 2 = 49.2
C base p = R + C base v = 1.987 + 49.2 = 51.187
Q = nC base pdT = 1/24.6 * 51.187 * 1 = 2.08 Cal.
3 years ago
pa1
357 Points
										
 
Sol. C base P/C base V = 7.6, n = 1 mole, ∆T = 50K(a) Keeping the pressure constant, dQ = du + dw,∆T = 50 K, γ = 7/6, m = 1 mole,dQ = du + dw ⇒ nC base VdT = du + RdT ⇒ du = nCpdT – RdT= 1 * Rγ/γ – 1 * dT – RdT = R *7/6/7/6-1 dT – RdT= DT – RdT = 7RdT = 6 RdT = 6 * 8.3 * 50 = 2490 J.(b) Kipping Volume constant, dv = nC base VdT= 1 * R/γ – 1 * dT = 1 * 8.3/7/6-1 * 50= 8.3 * 50 * 6 = 2490 J(c)Adiabetically dQ = 0,= [n * R/γ – 1(T base 1 – T base 2)] = 1 * 8.3/7/6 -1 (T base 1 – T base 2)] = 8.3 * 50 * 6 = 2490 J
2 years ago
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