The amount of heat energy required to raise the temperature of 1g of Helium at NTP, from $T_1$K to $T_2$K is
$A)\;\dfrac{3}{8}{N_a}{k_B}\left( {{T_2} - {T_1}} \right)$
$B)\;\dfrac{3}{2}{N_a}{k_B}\left( {{T_2} - {T_1}} \right)$
$C)\;\dfrac{3}{4}{N_a}{k_B}\left( {{T_2} - {T_1}} \right)$
$D)\;\dfrac{3}{4}{N_a}{k_B}\left( {\dfrac{{{T_2}}}{{{T_1}}}} \right)$
Answer
561.9k+ views
Hint: Here, the term, specific heat is required. So, the specific heat is that amount of heat that raises to the temperature of 1g of the substance by one degree Celsius or one Kelvin.
Complete step by step answer:
Here, the volume of the gas remains constant, thus the amount of the heat required to raise the temperature of the gas.
The expression is:
\[\Delta Q = n{C_V}\Delta T\] --- (1)
Here, $\Delta $Q is the change in heat.
$C_V$ is the Specific heat at constant volume
n is the number of moles,
$\Delta $T is the change in Temperature
So, we know that Helium is the monatomic molecule so, the value of specific heat at constant volume is
${C_V} = \dfrac{3}{2}R$
Here, R is the Universal Gas Constant
Number of moles, n = $\dfrac{1}{4}$
Change in temperature,$\Delta $T = ($T_2$ - $T_1$ )
So, no we put the value of change in temperature, number of moles and the specific heat at constant volume in equation (1)
$\Delta Q = \dfrac{1}{4} \times \dfrac{3}{2} \times R \times \left( {{T_2} - {T_1}} \right)$
$\Delta Q = \dfrac{3}{8} \times R \times \left( {{T_2} - {T_1}} \right)$ -- (2)
Here, the given option is in the form of the Boltzmann constant, $k_B$ and in the Avogadro constant $N_a$
The product of the Boltzmann constant, kB and the Avogadro constant, Na is equal to the molar gas constant.
$R = {k_B}{N_a}$
So, we put the value of R in the equation (2)
$ \Rightarrow \Delta Q = \dfrac{3}{8} \times {k_B} \times {N_a} \times \left( {{T_2} - {T_1}} \right)$
So, the change in heat is $\dfrac{3}{8}{N_a}{k_B}\left( {{T_2} - {T_1}} \right)$
So, the option (A) is correct.
Note:
Here, the difference between the universal gas constant and gas constant is that universal gas constant is only significant for the ideal gas; instead the gas constant is significant for the real gas.
Complete step by step answer:
Here, the volume of the gas remains constant, thus the amount of the heat required to raise the temperature of the gas.
The expression is:
\[\Delta Q = n{C_V}\Delta T\] --- (1)
Here, $\Delta $Q is the change in heat.
$C_V$ is the Specific heat at constant volume
n is the number of moles,
$\Delta $T is the change in Temperature
So, we know that Helium is the monatomic molecule so, the value of specific heat at constant volume is
${C_V} = \dfrac{3}{2}R$
Here, R is the Universal Gas Constant
Number of moles, n = $\dfrac{1}{4}$
Change in temperature,$\Delta $T = ($T_2$ - $T_1$ )
So, no we put the value of change in temperature, number of moles and the specific heat at constant volume in equation (1)
$\Delta Q = \dfrac{1}{4} \times \dfrac{3}{2} \times R \times \left( {{T_2} - {T_1}} \right)$
$\Delta Q = \dfrac{3}{8} \times R \times \left( {{T_2} - {T_1}} \right)$ -- (2)
Here, the given option is in the form of the Boltzmann constant, $k_B$ and in the Avogadro constant $N_a$
The product of the Boltzmann constant, kB and the Avogadro constant, Na is equal to the molar gas constant.
$R = {k_B}{N_a}$
So, we put the value of R in the equation (2)
$ \Rightarrow \Delta Q = \dfrac{3}{8} \times {k_B} \times {N_a} \times \left( {{T_2} - {T_1}} \right)$
So, the change in heat is $\dfrac{3}{8}{N_a}{k_B}\left( {{T_2} - {T_1}} \right)$
So, the option (A) is correct.
Note:
Here, the difference between the universal gas constant and gas constant is that universal gas constant is only significant for the ideal gas; instead the gas constant is significant for the real gas.
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