A gas mixture contains 1 g H2 and 1 g He- If temperature of gas mixture is increased from 0-C to 100-C at isobaric process- then find the heat given to the gas mixture- - -He - 53- -He - 75- R - 2 cal-mol-K-

Q = n_mix C_P_mixΔ T n_mix = n_1 + n_2 = 12 + 14 = 34 C_P_mix = n_1 C_P_1 + n_2 C_P_2n_1 + n_2 = 12. 7R2 + 14.5R212 + 14 = 196 R Q = (34 ) (196 ) (100 0)R = ( ...

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Byju's Answer

Standard XII

Physics

Specific Heat Capacity

A gas mixture...

Question

A gas mixture contains $1 g H_{2}$ and $1 g He$ . If temperature of gas mixture is increased from $0^{\circ} C$ to $100^{\circ} C$ at isobaric process, then find the heat given to the gas mixture. $[γ_{H e} = \frac{5}{3}, γ_{H e} = \frac{7}{5}, R = 2 cal/mol-K]$

124 cal

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327 cal

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218 cal

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475 cal

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Solution

The correct option is D $475 cal$
$Q = n_{m i x} C_{P_{m i x}} Δ T$
$n_{m i x} = n_{1} + n_{2} = \frac{1}{2} + \frac{1}{4} = \frac{3}{4}$
$C_{P_{m i x}} = \frac{n_{1} C_{P_{1}} + n_{2} C_{P_{2}}}{n_{1} + n_{2}} = \frac{\frac{1}{2} . \frac{7 R}{2} + \frac{1}{4} . \frac{5 R}{2}}{\frac{1}{2} + \frac{1}{4}} = \frac{19}{6} R$
$Q = (\frac{3}{4}) (\frac{19}{6}) (100 - 0) R = (19) (12.5) (2) = 475 cal$

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A gas mixture contains 1 g H2 and 1 g He. If temperature of gas mixture is increased from 0∘C to 100∘C at isobaric process, then find the heat given to the gas mixture. [γHe=53,γHe=75,R=2 cal/mol-K]

The correct option is D 475 calQ=nmixCPmixΔT nmix=n1+n2=12+14=34 CPmix=n1CP1+n2CP2n1+n2=12.7R2+14.5R212+14=196R Q=(34)(196)(100−0)R=(19)(12.5)(2)=475 cal

A gas mixture contains $1 g H_{2}$ and $1 g He$ . If temperature of gas mixture is increased from $0^{\circ} C$ to $100^{\circ} C$ at isobaric process, then find the heat given to the gas mixture. $[γ_{H e} = \frac{5}{3}, γ_{H e} = \frac{7}{5}, R = 2 cal/mol-K]$