A container contains 32 ,g of O_2 at a temper | vedclass

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

Question

(B) The ideal gas equation is given by $PV = nRT$,where $n = \frac{m}{M}$ is the number of moles.For the first container: $m(O_2) = 32 \,g$,$M(O_2) =

Vedclass · Accepted Answer

$4P$

Vedclass · Answer

(B) The ideal gas equation is given by $PV = nRT$,where $n = \frac{m}{M}$ is the number of moles.For the first container: $m(O_2) = 32 \,g$,$M(O_2) = 32 \,g/mol$. Thus,$n_1 = \frac{32}{32} = 1 \,mol$.The pressure is $P = \frac{n_1 RT}{V} = \frac{1 \cdot RT}{V} = \frac{RT}{V}$.For the second container: $m(H_2) = 4 \,g$,$M(H_2) = 2 \,g/mol$. Thus,$n_2 = \frac{4}{2} = 2 \,mol$.The temperature is $2T$. The pressure $P'$ is given by $P' = \frac{n_2 R(2T)}{V} = \frac{2 \cdot R(2T)}{V} = \frac{4RT}{V}$.Substituting $\frac{RT}{V} = P$,we get $P' = 4P$.

$A$ container contains $32 \,g$ of $O_2$ at a temperature $T$. The pressure of the gas is $P$. An identical container containing $4 \,g$ of $H_2$ at a temperature $2T$ has a pressure of ............

Similar Questions

From the following $P-T$ graph,what inference can be drawn?

The mass of oxygen gas occupying a volume of $11.2 \text{ litres}$ at a temperature $27^{\circ}C$ and a pressure of $760 \text{ mm}$ of mercury in kilograms is $[$ Molecular weight of oxygen $= 32]$

At a constant temperature,by what percentage $(\%)$ must the pressure of a given mass of gas be increased so that its volume decreases by $10\%$ (in $.1$)?

The adjoining figure shows the graph of pressure $(P)$ and volume $(V)$ of a gas at two temperatures $T_1$ and $T_2$. Which of the following inferences is correct?

$A$ flask containing air at $27\,^{\circ}C$ at atmospheric pressure is corked up. $A$ pressure of $2.5\,atm$ would force the cork out. The temperature at which it happens is

Vedclass Test Series

Exam Paper Generator

Online Exam Module