(A) Given initial conditions: $V_1 = 10 \,L$,$T_1 = 27^{\circ} C = 300 \,K$,$p_1 = 12 \,atm$.Final conditions: $V_2 = 6 \,L$,$T_2 = (27 + 30)^{\circ}

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(A) Given initial conditions: $V_1 = 10 \,L$,$T_1 = 27^{\circ} C = 300 \,K$,$p_1 = 12 \,atm$.Final conditions: $V_2 = 6 \,L$,$T_2 = (27 + 30)^{\circ} C = 57^{\circ} C = 330 \,K$.Using the ideal gas law,$\frac{p_1 V_1}{T_1} = \frac{p_2 V_2}{T_2}$.Rearranging for $p_2$: $p_2 = \frac{p_1 V_1 T_2}{T_1 V_2}$.Substituting the values: $p_2 = \frac{12 \times 10 \times 330}{300 \times 6}$.$p_2 = \frac{120 \times 330}{1800} = \frac{39600}{1800} = 22 \,atm$.

Class 11 Physics Kinetic Theory of Gases Gas Laws (Charles, Boyle's, Avagadro's, Gay Lussacs and Dalton's law) and Ideal gas Equation

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$A$ container of $10 \,L$ is filled with an ideal gas at a temperature of $27^{\circ} C$ at a pressure of $12 \,atm$. The volume of the container is reduced to $6 \,L$ and the temperature of the gas is increased by $30^{\circ} C$. What is the final pressure of the gas (in $\,atm$)?

AP EAMCET 2022 All AP EAMCET

A
$22$
B
$20$
C
$11$
D
$9$

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Kinetic Theory of Gases

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Gas Laws (Charles, Boyle's, Avagadro's, Gay Lussacs and Dalton's law) and Ideal gas Equation

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$A$ container of $10 \,L$ is filled with an ideal gas at a temperature of $27^{\circ} C$ at a pressure of $12 \,atm$. The volume of the container is reduced to $6 \,L$ and the temperature of the gas is increased by $30^{\circ} C$. What is the final pressure of the gas (in $\,atm$)?

Similar Questions

$A$ perfect gas at $27^{\circ}C$ is heated at constant pressure so as to double its volume. The final temperature of the gas will be,close to ...... $^{\circ}C$

When air is filled in a balloon, both the pressure and volume increase while the temperature remains constant. Why is Boyle's law not obeyed in this case?

At constant pressure,the ratio of the increase in volume of an ideal gas per degree rise in temperature to its original volume is ... (where $T$ is the absolute temperature of the gas).

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