Question

One litre of helium gas at a pressure of 76 cm-hg and temperature $ 27^\circ C $ is heated till its pressure and volume are doubled. The final temperature attained by the gas is
(A) $ 900^\circ C $
(B) $ 927^\circ C $
(C) $ 627^\circ C $
(D) $ 327^\circ C $

Answer 1

Hint: We need to assume that the gas is ideal. Compare the first state to the second state and make the temperature of the second state subject of the formula

Formula used: In this solution we will be using the following formulae;
 $ \dfrac{{{P_1}{V_1}}}{{{T_1}}} = \dfrac{{{P_2}{V_2}}}{{{T_2}}} $ where $ {P_1} $ is the pressure at the first state, $ {V_1} $ is the volume at the first state, and $ {T_1} $ is the temperature at the first state. Similarly, $ {P_2} $ is the pressure at the second state, $ {V_2} $ is the volume at the second state, and $ {T_2} $ is the temperature at that state.

Complete step by step solution:
To solve the above question, we assume that the helium gas is behaving like an ideal gas. Hence, we can use the ideal gas relation. From the ideal gas relation, we can get by comparison of the two state, the equation
 $ \dfrac{{{P_1}{V_1}}}{{{T_1}}} = \dfrac{{{P_2}{V_2}}}{{{T_2}}} $ where $ {P_1} $ is the pressure at the first state, $ {V_1} $ is the volume at the first state, $ {T_1} $ is the temperature at the first state, $ {P_2} $ is the pressure at the second state, $ {V_2} $ is the volume at the second state, and $ {T_2} $ is the temperature at that state.
Hence, by inserting all given values, we have
 $ \dfrac{{76\left( 1 \right)}}{{27 + 273}} = \dfrac{{\left( {2 \times 76} \right)\left( {2 \times 1} \right)}}{{{T_2}}} $ ( $ 27 + 273 $ since temperature has to be in kelvin)
Hence, $ {T_2} = \dfrac{{\left( {2 \times 76} \right)\left( {2 \times 1} \right) \times 300}}{{76}} $
Which by computation gives
 $ {T_2} = 1200K $
Which by conversion to degree Celsius is
 $ {T_C} = 1200 - 273 = 927^\circ C $
Hence, the correct answer is B.

Note:
Alternatively, observe that we do not need to insert the values for pressure and volume since we were told that it was doubled. We could simply write the equation as
 $ \dfrac{{{P_1}{V_1}}}{{{T_1}}} = \dfrac{{{P_2}{V_2}}}{{{T_2}}} = \dfrac{{\left( {2{P_1}} \right)\left( {2{V_1}} \right)}}{{{T_2}}} $
Hence, the pressure and volume values cancel out. So we have
 $ \dfrac{{1 \times 1}}{{{T_1}}} = \dfrac{{2 \times 2}}{{{T_2}}} $
 $ \Rightarrow {T_2} = 4 \times {T_1} $
From which we get
 $ {T_2} = 4 \times 300 = 1200K $ .