1 mole of rigid diatomic gas performs a work of $\frac{Q}{5}$ when heat Q is supplied to it. The molar heat capacity of the gas during this transformation is $\frac{x R}{8} .$ The value of x is

[R = universal gas constant]

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Payal Gupta | Contributor-Level 10

7 months ago

$Δ Q = Δ U + Δ W$

$\Rightarrow Q = Δ U + \frac{Q}{5} \Rightarrow Δ U = \frac{4 Q}{5} = n C_{v} Δ T \Rightarrow \frac{4 Q}{5} = \frac{5 R}{2} Δ T \Rightarrow Δ T = \frac{8 Q}{25 R}$

$Q = n c Δ T = 1 * C * \frac{80}{25 R} \Rightarrow C = \frac{25 R}{8} \Rightarrow x = 25$

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from newtan’s law of cooling :

$\frac{d T}{d t} = - k (T - T_{0})$

Using average value :

$\frac{75 - 65}{5} = - k [\frac{(75 + 65) 2}{2} - 25]$

$\frac{(65 - T)}{5} = - k [\frac{(65 + T)}{2} - 25]$

$\Rightarrow \frac{10}{65 - T} = \frac{45}{\frac{65 + T}{2} - 25}$

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A Carnot engine has efficiency of 50%. If the temperature of sink is reduced by 40°C, its efficiency increases by 30%. The temperature of the source will be:

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Efficiency $η = 50 %$

$\Rightarrow η = (1 - \frac{T_{2}}{T_{1}}) \times 100 = 50$

$\Rightarrow 1 - \frac{T_{2}}{T_{1}} = \frac{50}{100} \Rightarrow 1 - \frac{T_{2}}{T_{1}} = \frac{1}{2}$ -(i)

After the change :- $1 - \frac{T_{2}^{1}}{T_{1}} = \frac{50 \times (1.3)}{100} = \frac{65}{100} = . 65$ $\frac{_{}^{}}{_{}} \frac{}{} \frac{}{}$ -(ii)