Question

Forty calories of heat is needed to raise the temperature of 1mol of an ideal monatomic gas from $20{}^\circ C$ to $30{}^\circ C$ at a constant pressure. The amount of heat required to raise its temperature over the same interval at a constant volume ($R=2calmo{{l}^{-1}}{{K}^{-1}}$) is:
A. 20 cal
B. 40 cal
C. 60 cal
D. 80 cal

Answer 1

Hint: We are given the heat required to raise the temperature of 1 mol of an ideal gas from $20{}^\circ C$ to $30{}^\circ C$at constant pressure. And are asked to find the heat required to raise the temperature at the same level at constant volume. By using the equation of heat energy in an isobaric pressure we will get the specific heat at constant pressure and substituting that in the relation of specific heat at constant pressure and volume, we will get specific heat at constant volume. Using this we will get the heat at constant volume.
Formula used:
$\Delta q=n{{C}_{p}}\Delta T$
${{C}_{p}}-{{C}_{v}}=R$
$\Delta q=n{{C}_{v}}\Delta T$

Complete answer:
In the question it is said that 40 calories of heat is required to raise the temperature of 1 mol of an ideal monatomic gas from $20{}^\circ C$ to $30{}^\circ C$ at constant pressure.
Since the pressure is constant here, it is an isobaric process.
For an isobaric process we have change in heat energy is given as,
$\Delta q=n{{C}_{p}}\Delta T$, were ‘$\Delta q$’ is the change in heat energy, ‘$n$’ is the number of moles, ‘${{C}_{p}}$’ is the molar specific heat capacity of an ideal gas in an isobaric condition and ‘$\Delta T$’ is the change in temperature.
Here, in the question it is given that,
$\Delta q=1$
Number of moles, $n=1$
Change in temperature, $\Delta T=\left( 30-20 \right)=10$
Now we can substitute these values in the equation. Thus we get,
$\Rightarrow 40=1\times {{C}_{p}}\times 10$
By solving this equation we get the value of molar specific heat capacity of the given ideal gas at constant temperature as,
$\Rightarrow {{C}_{p}}=\dfrac{40}{10}=4calmo{{l}^{-1}}{{K}^{-1}}$
From the Mayer’s formula we know that,
${{C}_{p}}-{{C}_{v}}=R$, were ‘${{C}_{v}}$’ is the molar specific heat of an ideal gas at isochoric (constant volume) condition and ‘R’ is the ideal gas constant.
In the question we are given, $R=2calmo{{l}^{-1}}{{K}^{-1}}$ and from the earlier calculations we got ${{C}_{p}}=4calmo{{l}^{-1}}{{K}^{-1}}$.
By applying this in the Mayer’s formula, we get
$\Rightarrow 4-{{C}_{v}}=2$
By solving this we get molar specific heat at constant volume as,
$\Rightarrow {{C}_{v}}=\left( 4-2 \right)=2calmo{{l}^{-1}}{{K}^{-1}}$
We need to find the heat required to raise the temperature of the same gas over the same interval at constant volume.
We know that at constant volume heat is given by the equation,
$\Delta q=\Delta U=n{{C}_{v}}\Delta T$
Here,
$n=1$
${{C}_{v}}=2$
$\Delta T=\left( 30-20 \right)=10$
By substituting these values in the equation, we get the heat as,
$\Rightarrow \Delta q=1\times 2\times 10$
$\Rightarrow \Delta q=20cal$
Therefore the heat required to raise the temperature of the gas over the given interval at constant volume is 20 cal.

So, the correct answer is “Option A”.

Note:
The universal gas constant has various values in different units.
Here we take the value of R as $2calmo{{l}^{-1}}{{K}^{-1}}$ because this is the value of the gas constant in the terms of energy.
Similarly we take the value of gas constant based on the other parameters in the situation.