Two moles of monoatomic gas is mixed with three moles of a diatomic gas. The molar specific heat of the mixture at constant volume is

Question

Two moles of monoatomic gas is mixed with three moles of a diatomic gas. The molar specific heat of the mixture at constant volume is (A) 1.55R (B) 2.10R (C) 1.63R (D) 2.20R

Tardigrade Admin · Accepted Answer

Molar specific heat of the mixture at constant volume is (CV)mixture = (n1CV1+n2CV1/(n1+n2)) = (2((3/2)R)+3((5/2)R)/2+3) = 2.1R

Q. Two moles of monoatomic gas is mixed with three moles of a diatomic gas. The molar specific heat of the mixture at constant volume is

1.55R

2.10R

1.63R

2.20R

Molar specific heat of the mixture at constant volume is
$(C_{V})_{mi x t u re} = \frac{n _{1} C _{V_{1}} + n _{2} C _{V_{1}}}{( n _{1} + n _{2} )}$
$= \frac{2 ( \frac{3}{2} R ) + 3 ( \frac{5}{2} R )}{2 + 3} = 2.1 R$

Q. Two moles of monoatomic gas is mixed with three moles of a diatomic gas. The molar specific heat of the mixture at constant volume is

1.55R

2.10R

1.63R

2.20R

Molar specific heat of the mixture at constant volume is (CV​)mixture​=(n1​+n2​)n1​CV1​​+n2​CV1​​​ =2+32(23​R)+3(25​R)​=2.1R

Molar specific heat of the mixture at constant volume is
$(C_{V})_{mi x t u re} = \frac{n _{1} C _{V_{1}} + n _{2} C _{V_{1}}}{( n _{1} + n _{2} )}$
$= \frac{2 ( \frac{3}{2} R ) + 3 ( \frac{5}{2} R )}{2 + 3} = 2.1 R$