Question

A diatomic gas having ${C_P} = \dfrac{7}{2}R$ and ${C_V} = \dfrac{5}{2}R$, is heated at constant pressure. The ratio $dU:dQ:dW$ is
(A) $5:7:3$
(B) $5:7:2$
(C) $3:7:2$
(D) $3:5:2$

Answer 1

Hint: Here we will use the general formula of change in internal energy, change in heat energy, and change in work done on the gas, and using the given information we will solve for the asked ratio of $dU:dQ:dW$

Formula Used:
Change in internal energy of a gas is given by $dU = n{C_V}dT$ where, n is the number of moles of a gas, ${C_V}$ is known as specific heat at constant volume and dT is the temperature change.
Change in Heat energy is given by $dQ = n{C_P}dT$ where ${C_P}$ is called specific heat at constant pressure.
Change in work done is given by $dW = nRdT$ where R is the universal gas constant.

Complete answer:
We have given that for a diatomic gas the specific heat at constant volume is ${C_V} = \dfrac{5}{2}R$ and specific heat at constant pressure is ${C_P} = \dfrac{7}{2}R$

So, change in internal energy is calculated as $dU = n{C_V}dT$ we get,
$dU = n(\dfrac{7}{2})RdT \to (i)$

Change in heat energy is calculated using the formula $dQ = n{C_P}dT$ we get,
$dQ = n(\dfrac{5}{2})RdT \to (ii)$

Change in work done is calculated using the formula $dW = nRdT$ we get,
$dW = nRdT \to (iii)$

Now, take the ratio of equations (i), (ii), and (iii) together we get,
$dU:dQ:dW = 5:7:2$
So, the ratio is $dU:dQ:dW = 5:7:2$

Hence, the correct option is (B) $5:7:2$

Note: When we take the ratio for three values, always find the ratio of the first two values individually and then the ratio of the second and third value individually and then we write all ratios together, always make sure such calculations while solving such questions.