Correct answer is (1400)

W = nR$\mathrm{\Delta }$T = 400 J

$\therefore$ $\mathrm{\Delta }$Q = nC_P$\mathrm{\Delta }$T

$=n\times \frac{7}{2}R\times \mathrm{\Delta }T=\frac{7}{2}\times (400)=1400$

Correct answer is (2)

By 1^st law,

$\mathrm{\Delta }U=\mathrm{\Delta }Q-\frac{\mathrm{\Delta }Q}{4}=\frac{3}{4}\mathrm{\Delta }Q$

$\Rightarrow n{C}_v\mathrm{\Delta }T=\frac{3}{4}nC\mathrm{\Delta }T$

$\Rightarrow C=\frac{4C_v}{3}=2R$

Correct answer is (A)

$\mathrm{\Delta }$Q = $\mathrm{\Delta }$U + $\mathrm{\Delta }$W

$\frac{\mathrm{\Delta }Q}{\mathrm{\Delta }t}=\frac{\mathrm{\Delta }U}{\mathrm{\Delta }t}+\frac{\mathrm{\Delta }W}{\mathrm{\Delta }t}$

$\frac{6000}{60}\frac{J}{\sec }=\frac{2.5\times {10}^3}{\mathrm{\Delta }t}+90$

$\mathrm{\Delta }$t = 250 sec

Correct answer is (A)

According to first law of thermodynamics,

$\mathrm{\Delta }$Q = $\mathrm{\Delta }$U + $\mathrm{\Delta }$W ..... (i)

where, $\mathrm{\Delta }$Q = quantity of heat energy supplied to the system, $\mathrm{\Delta }$U = change in the internal energy of a closed system and $\mathrm{\Delta }$W = work done by the system on its surroundings.

As per question, no work is done

$\therefore$ $\mathrm{\Delta }$W = 0 ..... (iii)

From Eqs. (i) and (ii), we get

$\mathrm{\Delta }$Q = 0 + $\mathrm{\Delta }$U $\Rightarrow$ $\mathrm{\Delta }$Q = $\mathrm{\Delta }$U

or $\mathrm{\Delta }$Q = $\mathrm{\Delta }$U = nC_V$\mathrm{\Delta }$T

where,

C_V = specific heat capacity at constant volume for diatomic gas = $\frac{5R}{2}$

$\mathrm{\Delta }$T = change in temperature = (50 $-$ 0) = 50${}^{\circ }$C

n = number of moles = 4

$\Rightarrow$ $\mathrm{\Delta }$Q = nC_V$\mathrm{\Delta }$T

= $4\times \frac{5R}{2}\times (50)$ = 500 R = 500 R

Correct answer is (A)

Heat and work are path function.

Heat and work depends on the path taken to reach the final state from initial state.