NEET-XII-Physics
24: Kinetic Theory of Gases
- #270.040 g of He is kept in a closed container initially at 100.0°C. The container is now heated. Neglecting the expansion of the container, calculate the temperature at which the internal energy is increased by 12 J.Ans : `` \begin{array}{l}\text{Here,}\\ m=0.040g\\ M=4g\\ n=\frac{0.040}{4}=0.01\\ {T}_{1}=\left(100+273\right)K=373K\\ \text{He is a monoatomic gas. Thus,}\\ {C}_{v}=3\times \left(\frac{1}{2}R\right)\\ \Rightarrow {C}_{v}=1.5\times 8.3=12.45\\ {\text{Let the initial internal energy be U}}_{1}\text{.}\\ {\text{Let the final internal energy be U}}_{2}\text{.}\\ {U}_{2}-{U}_{1}=n{C}_{v}({T}_{2}-{T}_{1})\\ \Rightarrow 0.01\times 12.45({T}_{2}-373)=12\\ \Rightarrow {T}_{2}=469K\\ {\text{The temperature in}}^{0}C\text{can be obtained as follows:}\end{array}``
`` {\,\mathrm{\,469\; -\; 273\; =\; 196\,}}^{0}C``
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