
Volume of a gas at $ t^\circ C $ in $ 20 $ litre when temperature of gas increased by $ 50^\circ C $ at constant pressure then volume of gas increased by $ 25\% $ then calculated value of it?
Answer
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Hint: Charles law states that the temperature of the gas is directly proportional to the volume of the gas, when the pressure of the gas is constant. This reaction is also known as isobaric reaction, as the pressure is constant.
Formula used:
Charles law: $ T \propto V $
Where, $ T $ is temperature of the gas and $ V $ is the volume of the gas.
Also, represented as:
$ \dfrac{{{T_1}}}{{{T_2}}} = \dfrac{{{V_1}}}{{{V_2}}} $
Where, $ {T_1} $ is the initial temperature of the gas, $ {T_2} $ is the final temperature of the gas, $ {V_1} $ is the initial volume of the gas and $ {V_2} $ is the final volume of the gas.
Complete step by step solution:
As given the temperature of gas increases at constant pressure. Then, the reaction is an isobaric reaction.
Now, first of all we will discuss about the Charles law:
Charles law: This law states that the temperature of the gas is directly proportional to the volume of the gas when the pressure is constant.
$ T \propto V $
Where, $ T $ is temperature of the gas and $ V $ is the volume of the gas.
Representation, $ \dfrac{{{T_1}}}{{{T_2}}} = \dfrac{{{V_1}}}{{{V_2}}} $
Given, initial volume of the gas, $ {V_i} = 20l $
Initial temperature of the gas, $ {T_i} = t^\circ C $
In second case, the temperature of gas increased by $ 50^\circ C $ :
Final temperature of the gas, $ {T_f} = (t + 50)^\circ C $
The volume of the gas is increased by $ 25\% $
The final volume of the gas, $ {V_f} = \left( {\dfrac{{25}}{{100}} \times 20} \right) + 20 = 5 + 20 = 25l $
Now, putting the values in $ \dfrac{{{T_1}}}{{{T_2}}} = \dfrac{{{V_1}}}{{{V_2}}} $
$ \Rightarrow \dfrac{t}{{t + 50}} = \dfrac{{20}}{{25}} $
$ \Rightarrow 25t = 20\left( {t + 50} \right) $ $ \Rightarrow 25t - 20t = 1000 $
$ \Rightarrow 5t = 1000 \Rightarrow t = \dfrac{{1000}}{5} = 200^\circ C $
Hence, the value of $ t $ is $ 200^\circ C $ .
Hence, the Volume of a gas at $ 200^\circ C $ in $ 20 $ litre when temperature of gas increased by $ 50^\circ C $ at constant pressure then volume of gas increased by $ 25\% $ .
Note:
When the pressure is constant, and temperature is directly proportional to volume, it is known as isobaric. Similarly, when the temperature is constant, and pressure is inversely proportional to volume, it is known as isothermic and when the volume is constant, and pressure is directly proportional to temperature, it is known as isochoric.
Formula used:
Charles law: $ T \propto V $
Where, $ T $ is temperature of the gas and $ V $ is the volume of the gas.
Also, represented as:
$ \dfrac{{{T_1}}}{{{T_2}}} = \dfrac{{{V_1}}}{{{V_2}}} $
Where, $ {T_1} $ is the initial temperature of the gas, $ {T_2} $ is the final temperature of the gas, $ {V_1} $ is the initial volume of the gas and $ {V_2} $ is the final volume of the gas.
Complete step by step solution:
As given the temperature of gas increases at constant pressure. Then, the reaction is an isobaric reaction.
Now, first of all we will discuss about the Charles law:
Charles law: This law states that the temperature of the gas is directly proportional to the volume of the gas when the pressure is constant.
$ T \propto V $
Where, $ T $ is temperature of the gas and $ V $ is the volume of the gas.
Representation, $ \dfrac{{{T_1}}}{{{T_2}}} = \dfrac{{{V_1}}}{{{V_2}}} $
Given, initial volume of the gas, $ {V_i} = 20l $
Initial temperature of the gas, $ {T_i} = t^\circ C $
In second case, the temperature of gas increased by $ 50^\circ C $ :
Final temperature of the gas, $ {T_f} = (t + 50)^\circ C $
The volume of the gas is increased by $ 25\% $
The final volume of the gas, $ {V_f} = \left( {\dfrac{{25}}{{100}} \times 20} \right) + 20 = 5 + 20 = 25l $
Now, putting the values in $ \dfrac{{{T_1}}}{{{T_2}}} = \dfrac{{{V_1}}}{{{V_2}}} $
$ \Rightarrow \dfrac{t}{{t + 50}} = \dfrac{{20}}{{25}} $
$ \Rightarrow 25t = 20\left( {t + 50} \right) $ $ \Rightarrow 25t - 20t = 1000 $
$ \Rightarrow 5t = 1000 \Rightarrow t = \dfrac{{1000}}{5} = 200^\circ C $
Hence, the value of $ t $ is $ 200^\circ C $ .
Hence, the Volume of a gas at $ 200^\circ C $ in $ 20 $ litre when temperature of gas increased by $ 50^\circ C $ at constant pressure then volume of gas increased by $ 25\% $ .
Note:
When the pressure is constant, and temperature is directly proportional to volume, it is known as isobaric. Similarly, when the temperature is constant, and pressure is inversely proportional to volume, it is known as isothermic and when the volume is constant, and pressure is directly proportional to temperature, it is known as isochoric.
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