JEE MAIN - Physics (2021 - 20th July Morning Shift - No. 6)
The amount of heat needed to raise the temperature of 4 moles of rigid diatomic gas from 0$$^\circ$$ C to 50$$^\circ$$ C when no work is done is ___________. (R is the universal gas constant).
500 R
250 R
750 R
175 R
Explanation
According to first law of thermodynamics,
$$\Delta$$Q = $$\Delta$$U + $$\Delta$$W ..... (i)
where, $$\Delta$$Q = quantity of heat energy supplied to the system, $$\Delta$$U = change in the internal energy of a closed system and $$\Delta$$W = work done by the system on its surroundings.
As per question, no work is done
$$\therefore$$ $$\Delta$$W = 0 ..... (iii)
From Eqs. (i) and (ii), we get
$$\Delta$$Q = 0 + $$\Delta$$U $$\Rightarrow$$ $$\Delta$$Q = $$\Delta$$U
or $$\Delta$$Q = $$\Delta$$U = nCV$$\Delta$$T
where,
CV = specific heat capacity at constant volume for diatomic gas = $${{5R} \over 2}$$
$$\Delta$$T = change in temperature = (50 $$-$$ 0) = 50$$^\circ$$C
n = number of moles = 4
$$\Rightarrow$$ $$\Delta$$Q = nCV$$\Delta$$T
= $$4 \times {{5R} \over 2} \times (50)$$ = 500 R = 500 R
$$\Delta$$Q = $$\Delta$$U + $$\Delta$$W ..... (i)
where, $$\Delta$$Q = quantity of heat energy supplied to the system, $$\Delta$$U = change in the internal energy of a closed system and $$\Delta$$W = work done by the system on its surroundings.
As per question, no work is done
$$\therefore$$ $$\Delta$$W = 0 ..... (iii)
From Eqs. (i) and (ii), we get
$$\Delta$$Q = 0 + $$\Delta$$U $$\Rightarrow$$ $$\Delta$$Q = $$\Delta$$U
or $$\Delta$$Q = $$\Delta$$U = nCV$$\Delta$$T
where,
CV = specific heat capacity at constant volume for diatomic gas = $${{5R} \over 2}$$
$$\Delta$$T = change in temperature = (50 $$-$$ 0) = 50$$^\circ$$C
n = number of moles = 4
$$\Rightarrow$$ $$\Delta$$Q = nCV$$\Delta$$T
= $$4 \times {{5R} \over 2} \times (50)$$ = 500 R = 500 R
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