Question

One mole of helium is heated at ${0^ \circ }{\text{C}}$ and constant pressure. How much heat is required to increase the four-fold?
(A) $4095\;{\text{cal}}$.
(B) $5500\;{\text{cal}}$.
(C) $6000\;{\text{cal}}$.
(D) $6500\;{\text{cal}}$.

Answer 1

Hint In this question we will use the concept of the Charles law. First, we will calculate the change in temperature by using the Charles law which means that volume of the gas is directly proportional at constant pressure.

Complete step by step answer
In this question, we are given that one mole of helium is heated at constant pressure and the temperature is ${0^ \circ }{\text{C}}$.
Let represent the mole by $n$. One mole is,
$n = 1$
One should know the Charles law; Charles law says that volume of an ideal gas is directly proportional to the absolute temperature to constant pressure.
When the pressure is constant,
$\Rightarrow \dfrac{{{V_1}}}{{{T_1}}} = \dfrac{{{V_2}}}{{{T_2}}}$
Let us assume $\Delta T$ is the change in temperature, ${T_f}$ is the final temperature, and ${T_i}$ is the initial temperature.
We are given
${T_i} = 273\;{\text{k}}$
So, for fourfold increase in the temperature, the temperature is,
${T_i} = 4 \times 273\;{\text{k}}$
Now we will simplify the above expression.
$\Rightarrow {T_i} = 1092\;{\text{k}}$
Now we will calculate the change in temperature,
$\Delta T = {T_f} - {T_i}$
Now we substitute the value of change in temperature,
$\Rightarrow \Delta T = 1092 - 273$
Now we will simplify the above expression.
$\Rightarrow \Delta T = 819\;{\text{k}}$
As we know the valency of helium is $2$, so the helium is the diatomic molecule.
$\Rightarrow {C_V} = \dfrac{R}{{\gamma - 1}}$
Now we substitute the value
$\Rightarrow {C_V} = \dfrac{R}{{\dfrac{5}{3} - 1}}$
Further solving, we get
${C_V} = \dfrac{R}{{\dfrac{{5 - 3}}{3}}}$
$\Rightarrow {C_V} = \dfrac{3}{2}R$
Now we will calculate the ${C_p}$,
$\Rightarrow {C_P} = R + {C_V}$
For the monatomic gas, the value of ${C_p}$ is $\dfrac{5}{2}R$.
Now to calculate the change in heat using formula,
$\Delta Q = n{c_P}\Delta T$
Now we will substitute the values, where, $\Delta Q$ is the change in heat.
$\Rightarrow \Delta Q = 1 \times \dfrac{5}{2} \times 2 \times 819$
Further solving, we get
$\therefore \Delta Q = 4095\;{\text{cal}}$

Therefore, the option (A) is correct.

Note
In this question, do not forget to write the change in heat Si unit. The amount of heat is measured in calories. The calorie is the amount of the energy required for one degree Celsius to raise one gram of water.