Heat ;work and internal energy

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Question

A container closed by a freely moving piston contains a mixture of hydrogen and helium gases of total mass 12 gm. A heat of 150 kJ is transferred to the gas at constant pressure and in this process gas performs 50 kJ work. Find mass of hydrogen (in gm) in the mixture.

Moderate

Solution

$\begin{array}{l} {nC}_{P} ΔT = 150 \\ nRΔT = 50 \end{array}$
$\Rightarrow n C_{P} △ T = 3 n R △ T$
$\begin{array}{l} \Rightarrow n_{1} C_{P_{1}} + n_{2} C_{P_{2}} = 3 (n_{1} + n_{2}) R \\ \Rightarrow n_{1} (\frac{5 R}{2}) + n_{2} (\frac{7 R}{2}) = (3) (n_{1} + n_{2}) R \\ \Rightarrow 5 n_{1} + 7 n_{2} = 6 n_{1} + 6 n_{2} \\ \Rightarrow n_{1} = n_{2} \\ A l s o, n_{1} (4) + n_{2} (2) = 12 \\ n_{1} = n_{2} = 2 \end{array}$
Mass of hydrogen = 4 gm

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Heat ;work and internal energy

Remember concepts with our Masterclasses.

A container closed by a freely moving piston contains a mixture of hydrogen and helium gases of total mass 12 gm. A heat of 150 kJ is transferred to the gas at constant pressure and in this process gas performs 50 kJ work. Find mass of hydrogen (in gm) in the mixture.

nCPΔT=150nRΔT=50 ⇒nCP△T=3nR△T⇒n1CP1+n2CP2=3n1+n2R⇒n15R2+n27R2=(3)n1+n2R⇒5n1+7n2=6n1+6n2⇒n1=n2Also, n1(4)+n2(2)=12n1=n2=2Mass of hydrogen = 4 gm

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