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QPages 77-80: Q14: HC Verma: Concepts of Physics - Volume 2 Specific Heat Capacities of Gases Pages 77-80: Q14 Solution: An ideal gas ) is taken through the process abc shown in the…

An ideal gas γ=1.67\gamma = 1.67) is taken through the process abc shown in the figure. The temperature at point a is 300 K. Calculate a) the temperatures at b and c b) the work done in the process c) the amount of heat supplied in the path ab and in the path bc and d) the change in the internal energy of the gas in the process.

Step-by-Step Verified Solution

**Answer:** (a) Tb=600T_b = 600 K, Tc=900T_c = 900 K (b) W=10W = 10 J (c) Qab=14.925Q_{ab} = 14.925 J, Qbc=24.925Q_{bc} = 24.925 J (d) ΔU=29.85\Delta U = 29.85 J. **Step-by-Step Explanation:** Given: Ta=300T_a = 300 K, γ=1.67\gamma = 1.67. (a) For abab (constant volume): PaTa=PbTb⇒Tb=600\frac{P_a}{T_a} = \frac{P_b}{T_b} \Rightarrow T_b = 600 K. For bcbc (constant pressure): VbTb=VcTc⇒Tc=900\frac{V_b}{T_b} = \frac{V_c}{T_c} \Rightarrow T_c = 900 K. (b) W=PΔV=200×103×50×10−6=10W = P\Delta V = 200 \times 10^3 \times 50 \times 10^{-6} = 10 J. (c) Qab=nCvΔT=14.925Q_{ab} = nC_v\Delta T = 14.925 J, Qbc=nCpΔT=24.925Q_{bc} = nC_p\Delta T = 24.925 J. (d) ΔU=Qtotal−W=(14.925+24.925)−10=29.85\Delta U = Q_{total} - W = (14.925 + 24.925) - 10 = 29.85 J [citation:12].

All Questions in this Chapter

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