Question

A mixture contains 1 mole of helium (Cp = 2.5 R, Cv = 1.5 R) and 1 mole of hydrogen (Cp = 3.5 R, Cv = 2.5 R). Calculate the values of Cp, Cv and γ for the mixture.

Answer 1

Answer:

Given,

CP1 = 2.5 R, CV1= 1.5R for helium

CP2 = 3.5 R, CV2= 2.5 R for hydrogen

n1=n2=1

We know dU =nCvdT

Where dU is the change in internal energy, n is the number of moles, Cv is the molar heat capacity at constant volume and dT is the change in temperature.

For the mixture,







Also,





Answer 2

$C_{P}=$$3 R$, $C_{v}=$$2 R$, $\gamma$$=1.5$

Explanation:

Step 1:

Specific heat, at constant helium pressure, $C_{p}^{\prime}=2.5 R$

Specific heat, at constant hydrogen pressure, $C_{p}^{\prime \prime}=3.5 R$

Specific heat at constant helium volume, $C_{\mathrm{v}}^{\prime}=1.5 R$

Specific gas, with constant hydrogen volume, $C_{V}^{\prime \prime}=2.5 R$

$n_{1}=n_{2}=1 \mathrm{mol}$

$\mathrm{d} \mathrm{U}=\mathrm{nC}_{\mathrm{v}} \mathrm{dT}$

Step 2:

For the mixing of two gases,

$\mathrm{d} \mathrm{U}_{1}+\mathrm{d} \mathrm{U}_{2}=1 \mathrm{mol}$

$\left[n_{1}+n_{2}\right] C_{v} d T=n_{1} C_{v}^{\prime} d T+n_{2} C_{v}^{\prime \prime} d T$

Step 3:

Where CV is the compound heat capacity.

$C_{v}=\frac{n_{1} C^{\prime} v+n_{2} C^{\prime \prime} v}{n_{1}+n_{2}}$

$=\frac{1.5 R+2.5 R}{2}$

$=\frac{4 R}{2}=2 R$

Step 4:

$C_{P}=C_{v}+\mathrm{R}=2 \mathrm{R}+\mathrm{R}=3 \mathrm{R}$

$\gamma=\frac{C_{P}}{C_{v}}=\frac{3 R}{2 R}=1.5$