Question

A mixture of 8gm of helium and 14gm of nitrogen is enclosed in a vessel of constant volume at 300K. The quantity of heat absorbed by the mixture to double the root mean velocity of its molecules is –(R = universal gas constant )

• A. 2725 R
• B. 3630 R
• C. 3825 R
• D. 5625 R

Answer 1

Answer: (C)

3825 R

Answer 2

Q = DU + W

W = 0 since volume is constant

Q = DU

V_rms = $\sqrt { 3 \mathrm { RT } / \mathrm { M } }$

U_mix =U₁ + U₂

U_mix = n₁T + n₂ $\mathrm { C } _ { \mathrm { v } _ { 2 } }$ T = (n₁ + n₂) (C_v)_mix T

(U_f) – (U_i) = nC_v(T₂ – T₁) = (n₁ + n₂) (C_v)_mix (T₂–T₁)

V¢_rms = 2V_rms Ž T¢ = 4T

n₁ = $\frac { 8 } { 4 }$ = 2 ; n₂ = $\frac { 14 } { 28 }$ = 1/2

(C_v)mix = $\frac { \mathrm { n } _ { 1 } \mathrm { Cv } _ { 1 } + \mathrm { n } _ { 2 } \mathrm { Cv } _ { 2 } } { \left( \mathrm { n } _ { 1 } + \mathrm { n } _ { 2 } \right) }$

U_f – U_i = (n₁ $\mathrm { C } _ { \mathrm { v } _ { 1 } }$ + n₂) (T₂ – T₁)

=$\left( 2 \times \frac { 3 } { 2 } + \frac { 1 } { 2 } \times \frac { 5 } { 2 } \right)$ R [1200 – 300]

$\frac { 17 } { 4 }$ × R × 100 = 3825 R