Chapter 8: Problem 1118
If heat given to a system is \(6 \mathrm{k}\) cal and work done is $6 \mathrm{kj}$. The change in internal energy is ......... KJ. (A) \(12.4\) (B) 25 (C) \(19.1\) (D) 0
Chapter 8: Problem 1118
If heat given to a system is \(6 \mathrm{k}\) cal and work done is $6 \mathrm{kj}$. The change in internal energy is ......... KJ. (A) \(12.4\) (B) 25 (C) \(19.1\) (D) 0
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Get started for freeThe first law of thermodynamics is concerned with the conservation of (A) momentum (B) energy (C) mass (D) temperature
For an adiabatic process involving an ideal gas (A) \(\mathrm{P}^{\gamma-1}=\mathrm{T}^{\gamma-1}=\) constant (B) \(\mathrm{P}^{1-\gamma}=\mathrm{T}^{\gamma}=\) constant (C) \(\mathrm{PT}^{\gamma-1}=\) constant (D) \(\mathrm{P}^{\gamma-1}=\mathrm{T}^{\gamma}=\) constant
For adiabatic Process which relation is true mentioned below ? \(\gamma=\left\\{\mathrm{C}_{\mathrm{p}} / \mathrm{C}_{\mathrm{v}}\right\\}\) (A) \(\mathrm{p}^{\gamma} \mathrm{V}=\mathrm{Const}\) (B) \(\mathrm{T}^{\gamma} \mathrm{V}=\mathrm{Const}\) (C) TV \(^{\gamma}=\) Const (D) \(\mathrm{TV}^{\gamma-1}=\) Const
A System under goes a Cyclic Process in which it absorbs \(\mathrm{Q}_{1}\) heat and gives out \(\mathrm{Q}_{2}\) heat. The efficiency of the Process \(\eta\) and the work done is \(\mathrm{W}\). Which formula is wrong ? is (A) \(\mathrm{W}=\mathrm{Q}_{1}-\mathrm{Q}_{2}\) (B) \(\eta=\left(\mathrm{Q}_{2} / \mathrm{Q}_{1}\right)\) (C) \(\eta=\left(\mathrm{W} / \mathrm{Q}_{1}\right)\) (D) \(\eta=1-\left(\mathrm{Q}_{2} / \mathrm{Q}_{1}\right)\)
A Carnot engine takes \(3 \times 10^{6}\) cal of heat from a reservoir at \(627^{\circ} \mathrm{C}\), and gives to a sink at \(27^{\circ} \mathrm{C}\). The work done by the engine is (A) \(4.2 \times 10^{6} \mathrm{~J}\) (B) \(16.8 \times 10^{6} \mathrm{~J}\) (C) \(8.4 \times 10^{6} \mathrm{~J}\) (D) Zero
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