A mixture contains 1 mole of helium $(C_{P} = 2.5 R, C_{V} = 1.5 R)$ and 1 mole of hydrogen $(C_{P} = 3.5 R, C_{V} = 2.5 R)$ . If the value of $C_{V}$ for the mixture is xR then what is the value of x?

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Solution

Given : $n_{H e} = 1$ $C_{V_{H e}} = 1.5 R$ $n_{H_{2}} = 1$ $C_{V_{H_{2}}} = 2.5 R$
Value of $C_{V}$ for mixture $C_{V} = \frac{n_{H e} C_{V_{H e}} + n_{H_{2}} C_{V_{H_{2}}}}{n_{H e} + n_{H_{2}}}$
$∴$ $C_{V} = \frac{(1) (1.5 R) + (1) (2.5 R)}{1 + 1} = 2 R$
Thus we get $x = 2$

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A mixture contains 1 mole of helium (CP=2.5R,CV=1.5R) and 1 mole of hydrogen (CP=3.5R,CV=2.5R). If the value of CV for the mixture is xR then what is the value of x?

Given : nHe=1 CVHe=1.5R nH2=1 CVH2=2.5RValue of CV for mixture CV=nHeCVHe+nH2CVH2nHe+nH2∴ CV=(1)(1.5R)+(1)(2.5R)1+1=2RThus we get x=2

A mixture contains 1 mole of helium $(C_{P} = 2.5 R, C_{V} = 1.5 R)$ and 1 mole of hydrogen $(C_{P} = 3.5 R, C_{V} = 2.5 R)$ . If the value of $C_{V}$ for the mixture is xR then what is the value of x?

Given : $n_{H e} = 1$ $C_{V_{H e}} = 1.5 R$ $n_{H_{2}} = 1$ $C_{V_{H_{2}}} = 2.5 R$
Value of $C_{V}$ for mixture $C_{V} = \frac{n_{H e} C_{V_{H e}} + n_{H_{2}} C_{V_{H_{2}}}}{n_{H e} + n_{H_{2}}}$
$∴$ $C_{V} = \frac{(1) (1.5 R) + (1) (2.5 R)}{1 + 1} = 2 R$
Thus we get $x = 2$