
A certain amount of gas at ${27^o}C$ and one bar pressure occupied a volume of $25{m^3}$ . If pressure kept constant temperature raised to ${77^o}C$ , what will be the volume of gas?
Answer
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Hint:We already have three parameters in the question given and we have to find the fourth parameter . If we substitute them in the Charles law formula, we will be able to solve the given problem.
Law used:
The formula of Charles law is used here. It is the law based on the behavior of gases’ expansion during applied temperature and pressure. It states that the product of pressure and temperature of one gas is equal to the product of pressure and temperature of another gas in equilibrium. This law is used to find the partial pressures, temperatures of individual gases or others in a mixture of several gases.
Formula used:
${V_1}{T_1} = {V_2}{T_2}$
Where,
${V_1}$ - Volume of gas 1.
${V_2}$ - Volume of gas 2.
${T_1}$ - Temperature of gas 1.
${T_2}$ - Temperature of gas 2.
Complete step by step answer:
Let the gas of temperature ${27^o}C$ and 1 bar pressure be gas 1, and the other be gas 2.
${T_1} = {27^o}C = 300K$
${V_1} = 25{m^3}$
${T_2} = {77^o}C = 350K$
Substituting the values in the formula that we know,
${V_1}{T_1} = {V_2}{T_2}$
$25 \times 300 = {V_2} \times 350$
Taking the value of ${T_2}$ to the left hand side,
${V_2} = \dfrac{{25 \times 300}}{{350}}{m^3}$
${V_2} = 21.43{m^3}$
Therefore, the volume of the second gas will be $21.43{m^3}$.
ADDITIONAL INFORMATION:
The questions can be of any parameter based. They shall ask for ${T_2}$ , ${V_1}$ or ${V_2}$. Applying the similar steps though the parameters can give you the answers, since the answers are based on a single equation. Also, the same law can be applied for solving pressures of both gases when the given condition is in equilibrium and any one of the gases’ pressure is provided.
Note:Always check for the questions whether the mixture is in equilibrium or not. Since, this law suits best for equilibrium mixtures, continue using the Charles law. If pressure is given and have been mentioned about a single gas, use Gay Lussac’s law. Also, knowing the values of constants and units is a must .
Law used:
The formula of Charles law is used here. It is the law based on the behavior of gases’ expansion during applied temperature and pressure. It states that the product of pressure and temperature of one gas is equal to the product of pressure and temperature of another gas in equilibrium. This law is used to find the partial pressures, temperatures of individual gases or others in a mixture of several gases.
Formula used:
${V_1}{T_1} = {V_2}{T_2}$
Where,
${V_1}$ - Volume of gas 1.
${V_2}$ - Volume of gas 2.
${T_1}$ - Temperature of gas 1.
${T_2}$ - Temperature of gas 2.
Complete step by step answer:
Let the gas of temperature ${27^o}C$ and 1 bar pressure be gas 1, and the other be gas 2.
${T_1} = {27^o}C = 300K$
${V_1} = 25{m^3}$
${T_2} = {77^o}C = 350K$
Substituting the values in the formula that we know,
${V_1}{T_1} = {V_2}{T_2}$
$25 \times 300 = {V_2} \times 350$
Taking the value of ${T_2}$ to the left hand side,
${V_2} = \dfrac{{25 \times 300}}{{350}}{m^3}$
${V_2} = 21.43{m^3}$
Therefore, the volume of the second gas will be $21.43{m^3}$.
ADDITIONAL INFORMATION:
The questions can be of any parameter based. They shall ask for ${T_2}$ , ${V_1}$ or ${V_2}$. Applying the similar steps though the parameters can give you the answers, since the answers are based on a single equation. Also, the same law can be applied for solving pressures of both gases when the given condition is in equilibrium and any one of the gases’ pressure is provided.
Note:Always check for the questions whether the mixture is in equilibrium or not. Since, this law suits best for equilibrium mixtures, continue using the Charles law. If pressure is given and have been mentioned about a single gas, use Gay Lussac’s law. Also, knowing the values of constants and units is a must .
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