At what temperature will the volume of a gas at 0 c double itself

Answer

Verified

Hint : As we know very well that the behaviours of the gases are described by the gas equation. Especially in these types of questions, we generally assume the gas to be the ideal gas, then we can explain the answer easily by applying the ideal gas law or ideal gas equation.

Complete Step By Step Answer:

So first we will describe the ideal gas law and its equation:So, let’s assume that the given gas is an ideal gas and all the laws of the ideal gas equation will be applicable to it.So, let’s see the ideal gas equation and it is: $ \Rightarrow $ $ PV = nRT $ Where, P is the pressure of the gasV is the volume of the gas.N is the number of molesR is the gas constantAnd T is the temperature of the gas.So, according to the conditions given in the question i.e. constant pressure.So, we can say that, At constant pressure P, and with a fixed number of moles n, the volume of the gas is directly proportional to the temperature. $ V\alpha T $ Keep in mind here that R is the gas constant.So, we can say as the volume increases or gets doubled, the temperature will also increase or will get doubled.Hence, our equation will be: $ \Rightarrow $ $ \dfrac{{{V_1}}}{{{T_1}}} = \dfrac{{{V_2}}}{{{T_2}}} $ Here, our values are: $ {V_1} = V \\ {T_1} = 273K \\ {V_2} = 2V \\ {T_2} = ? \\ $ Putting these values in the formula, we will get: $ \dfrac{V}{{273}} = \dfrac{{2V}}{{{T_2}}} \\ {T_2} = 273 \times 2 \\ {T_2} = 546K \\ $ Converting this temperature from kelvin to the degree Celsius scale, we will get: $ T(K) = T{(^O}C) + 273 \\ $ \Rightarrow $ T{(^O}C) = 273{{\text{ }}^O}C \\ $

Therefore the correct answer is option B i.e. $ 273{{\text{ }}^O}C $ .

Note :

As in the question above, we derived the formula of the relationship between the volume and temperature from the gas law, but we can also directly use this relation as given by Charles law. According to Charles law, the volume of the ideal gas is directly proportion to the temperature.

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Text Solution

Answer : The required temperature is `546^@C`.

Solution : Let the volume at `0^@C` = V ml. `V_1 = VmL, V_2 = 3 mL` `T_1 = 0^@` or `273 K, T_2 = ?` by charles law, ` V_1/T_1 = V_2/T_2` or `V/273 =(3V)/T_2` or `T_2 = (3xx273) = 819 K` or `T_2 = (819-273)^@C = 546^@C`

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