Two moles of an ideal gas are heated at a constant pressure of 1 atm from 27-C to 127-C- If Cv-m -20 - 10-2 T JK-1mol-1- then q and - U for the process are respectively-

Q = ∫ nC_p.m^dT = ∫^400 K_300 K 2×(853 + 10^ 2 T ) dT = 1703(400 300) + 2 × 10^ 22(400^2 300^2) = 6366.66 J E = ∫ nC_v, m dT = ∫^400 K_300 K 2 × (20 + 10 ...

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Standard XI

Chemistry

First Law of Thermodynamics

Two moles of ...

Question

Two moles of an ideal gas are heated at a constant pressure of 1 atm from $27^{\circ} C to 127^{\circ} C$ . If $C_{v, m} = 20 + 10^{- 2} T J K^{- 1} m o l^{- 1}$ , then q and $Δ U$ for the process are respectively:

6366.7 J, 4700 J

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3033.3 J, 4700 J

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7066.7 J, 5400 J

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4016.7 J, 2350 J

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Solution

The correct option is A 6366.7 J, 4700 J
$q = \int n C_{p . m}^{d T} = \int_{300 K}^{400 K} 2 \times (\frac{85}{3} + 10^{- 2} T) d T$
$= \frac{170}{3} (400 - 300) + \frac{2 \times 10^{- 2}}{2} (400^{2} - 300^{2})$
$= 6366.66 J$
$△ E = \int n C_{v, m} d T = \int_{300 K}^{400 K} 2 \times (20 + 10^{- 2} T) d T$
$= 40 \times (400 - 300) + \frac{2 \times 10^{- 2}}{2} (400^{2} - 300^{2})$
$= 4700 J$

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Two moles of an ideal gas are heated at a constant pressure of 1 atm from 27∘C to 127∘C. If Cv,m=20+10−2 TJK−1mol−1, then q and ΔU for the process are respectively:

The correct option is A 6366.7 J, 4700 Jq=∫nCdTp.m=∫400 K300 K2×(853+10−2T)dT =1703(400−300)+2×10−22(4002−3002) =6366.66 J △E=∫nCv,mdT=∫400 K300 K2×(20+10−2T)dT =40×(400−300)+2×10−22(4002−3002) =4700 J

Two moles of an ideal gas are heated at a constant pressure of 1 atm from $27^{\circ} C to 127^{\circ} C$ . If $C_{v, m} = 20 + 10^{- 2} T J K^{- 1} m o l^{- 1}$ , then q and $Δ U$ for the process are respectively: