22 g of carbon dioxide at 27^{circ} C is mix | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) Let $T^{\circ} C$ be the final equilibrium temperature of the mixture.Since the container is closed and isolated,the heat lost by the hotter gas $

Vedclass · Accepted Answer

$31.5$

Vedclass · Answer

(C) Let $T^{\circ} C$ be the final equilibrium temperature of the mixture.Since the container is closed and isolated,the heat lost by the hotter gas $(O_2)$ equals the heat gained by the cooler gas $(CO_2)$.For $CO_2$ (a non-linear triatomic gas),the molar heat capacity at constant volume is $C_{V, CO_2} = 3R$.For $O_2$ (a diatomic gas),the molar heat capacity at constant volume is $C_{V, O_2} = \frac{5}{2}R$.The number of moles are: $\mu_{CO_2} = \frac{22}{44} = 0.5 \ mol$ and $\mu_{O_2} = \frac{16}{32} = 0.5 \ mol$.Applying the principle of calorimetry: $\mu_{CO_2} C_{V, CO_2} (T - 27) = \mu_{O_2} C_{V, O_2} (37 - T)$.$0.5 	imes 3R 	imes (T - 27) = 0.5 	imes \frac{5}{2}R 	imes (37 - T)$.$3(T - 27) = 2.5(37 - T)$.$3T - 81 = 92.5 - 2.5T$.$5.5T = 173.5$.$T = \frac{173.5}{5.5} \approx 31.5^{\circ} C$.

$22 \ g$ of carbon dioxide at $27^{\circ} C$ is mixed in a closed container with $16 \ g$ of oxygen at $37^{\circ} C$. If both gases are considered as ideal gases,then the temperature of the mixture is nearly: (in $^{\circ} C$)

Similar Questions

$A$ closed vessel contains $0.1$ mole of a monoatomic ideal gas at $200\, K$. If $0.05$ mole of the same gas at $400\, K$ is added to it,the final equilibrium temperature (in $K$) of the gas in the vessel will be closest to:

Two moles of an ideal gas with $\frac{C_{P}}{C_{V}}=\frac{5}{3}$ are mixed with $3$ moles of another ideal gas with $\frac{C_{P}}{C_{V}}=\frac{4}{3}$. The value of $\frac{C_{P}}{C_{V}}$ for the mixture is

Two gases $A$ and $B$ having same pressure $P$,volume $V$,and absolute temperature $T$ are mixed. If the mixture has the volume and temperature as $V$ and $T$ respectively,then the pressure of the mixture is

The pressure and temperature of two different gases are $P$ and $T$,each having the same volume $V$. If they are mixed while keeping the volume and temperature constant,the pressure of the mixture will be:

One litre of oxygen at a pressure of $1 \,atm$ and two litres of nitrogen at a pressure of $0.5 \,atm$ are introduced into a vessel of volume $1 \,L$. If there is no change in temperature, the final pressure of the mixture of gas (in $atm$) is

Vedclass Test Series

Exam Paper Generator

Online Exam Module