A gas mixture contains 1 g H 2 and 1 g He- If temperature of gas mixture is increased from 0- C to 100- C at isobaric process- then find the heat given to the gas mixture- - YHe -5-3- YHe -7-5- R -2 cal - mol - K -A- 124 calB- 218 calC- 327 calD- 475 cal

The correct option is D 475 calQ = n_mix C_P_mixΔ T n_mix = n_1 + n_2 = 12 + 14 = 34 C_P_mix = n_1 C_P_1 + n_2 C_P_2n_1 + n_2 = 12. 7R2 + 14.5R212 + 14 = 196 R ...

You visited us 1 times! Enjoying our articles? Unlock Full Access!

Byju's Answer

Standard XII

Physics

Specific Heat Capacity

A gas mixture...

Question

A gas mixture contains $1 g H_{2}$ and $1 g He$ . If temperature of gas mixture is increased from $0^{\circ} C$ to $100^{\circ} C$ at isobaric process, then find the heat given to the gas mixture. $[γ_{H e} = \frac{5}{3}, γ_{H e} = \frac{7}{5}, R = 2 cal/mol-K]$

124 cal

No worries! We‘ve got your back. Try BYJU‘S free classes today!

327 cal

No worries! We‘ve got your back. Try BYJU‘S free classes today!

218 cal

No worries! We‘ve got your back. Try BYJU‘S free classes today!

475 cal

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

Open in App

Solution

The correct option is D $475 cal$
$Q = n_{m i x} C_{P_{m i x}} Δ T$
$n_{m i x} = n_{1} + n_{2} = \frac{1}{2} + \frac{1}{4} = \frac{3}{4}$
$C_{P_{m i x}} = \frac{n_{1} C_{P_{1}} + n_{2} C_{P_{2}}}{n_{1} + n_{2}} = \frac{\frac{1}{2} . \frac{7 R}{2} + \frac{1}{4} . \frac{5 R}{2}}{\frac{1}{2} + \frac{1}{4}} = \frac{19}{6} R$
$Q = (\frac{3}{4}) (\frac{19}{6}) (100 - 0) R = (19) (12.5) (2) = 475 cal$

Suggest Corrections

Similar questions

Q. A gas mixture contain

1 g

H_{2}

and

1 g

H_{e}

temperature of gas mixture is increased from

0^{\circ}

100^{0} C

at isobaric process. Then find given heat of gas mixture

[γ_{H e} = 5 / 3, γ_{H e} = 7 / 5, R = 2 c a l / m o l - K]

Q. A gas mixture contains

1 g H_{2}

and

1 g He

. If temperature of gas mixture is increased from

0^{\circ} C

100^{\circ} C

at isobaric process, then find the heat given to the gas mixture.

[γ_{H e} = \frac{5}{3}, γ_{H e} = \frac{7}{5}, R = 2 cal/mol-K]

Q. How much heat energy should be added to the gaseous mixture consisting of

1 g

of hydrogen and

1 g

of helium to raise its temperature from

0^{\circ} C

100^{\circ} C

at constant volume,

(R = 2 c a l / m o l K)

A mixture of ideal gas containing 5 moles of monatomic gas and 1 mole of rigid diatomic gas is initially at pressure $P_{0}$ , volume $V_{0}$ and temperature $T_{0}$ . If the gas mixture is adiabatically compressed to a volume $V_{0} / 4$ , then the correct statement(s) is/are,
(Given $2^{1.2} = 2.3; 2^{3.2} = 9.2; R$ is gas constant)

Q. A gas mixture contains one mole

O_{2}

gas and one mole He gas. Find the ratio of specific heat at constant pressure to that at constant volume of the gaseous mixture.

Join BYJU'S Learning Program

A gas mixture contains 1 g H2 and 1 g He. If temperature of gas mixture is increased from 0∘C to 100∘C at isobaric process, then find the heat given to the gas mixture. [γHe=53,γHe=75,R=2 cal/mol-K]

The correct option is D 475 calQ=nmixCPmixΔT nmix=n1+n2=12+14=34 CPmix=n1CP1+n2CP2n1+n2=12.7R2+14.5R212+14=196R Q=(34)(196)(100−0)R=(19)(12.5)(2)=475 cal

A gas mixture contains $1 g H_{2}$ and $1 g He$ . If temperature of gas mixture is increased from $0^{\circ} C$ to $100^{\circ} C$ at isobaric process, then find the heat given to the gas mixture. $[γ_{H e} = \frac{5}{3}, γ_{H e} = \frac{7}{5}, R = 2 cal/mol-K]$