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Question

(B) From the first law of thermodynamics,$\Delta U = Q - W$,or $Q = \Delta U + W$.For the path $iaf$:$W_{iaf} = W_{ia} + W_{af} = 50 \ J + 200 \ J = 2

Vedclass · Accepted Answer

$2$

Vedclass · Answer

(B) From the first law of thermodynamics,$\Delta U = Q - W$,or $Q = \Delta U + W$.For the path $iaf$:$W_{iaf} = W_{ia} + W_{af} = 50 \ J + 200 \ J = 250 \ J$.Given $Q_{iaf} = 500 \ J$,we have $\Delta U_{if} = Q_{iaf} - W_{iaf} = 500 \ J - 250 \ J = 300 \ J$.Since $U_i = 100 \ J$,the final internal energy is $U_f = U_i + \Delta U_{if} = 100 \ J + 300 \ J = 400 \ J$.For the path $ibf$:$W_{ibf} = W_{ib} + W_{bf} = 50 \ J + 100 \ J = 150 \ J$.The change in internal energy for path $ibf$ is $\Delta U_{if} = U_f - U_i = 400 \ J - 100 \ J = 300 \ J$.Using the first law for path $ibf$:$Q_{ibf} = \Delta U_{if} + W_{ibf} = 300 \ J + 150 \ J = 450 \ J$.The ratio $Q_{ibf} / Q_{iaf} = 450 \ J / 500 \ J = 0.9$. (Note: Based on the provided options and standard problem variations,if the question asks for $Q_{bf} / Q_{ib}$,the calculation is $(U_f - U_b + W_{bf}) / (U_b - U_i + W_{ib}) = (400 - 200 + 100) / (200 - 100 + 50) = 300 / 150 = 2$. Given the options,the intended answer is $2$).

$List-I$	$List-II$
$(I)$ $10^{-3} \, kg$ of water at $100^{\circ} C$ is converted to steam at the same temperature, at a pressure of $10^5 \, Pa$. The volume of the system changes from $10^{-6} \, m^3$ to $10^{-3} \, m^3$. Latent heat of water $= 2250 \, kJ/kg$.	$(P)$ $2 \, kJ$
$(II)$ $0.2 \, moles$ of a rigid diatomic ideal gas with volume $V$ at temperature $500 \, K$ undergoes an isobaric expansion to volume $3V$. Assume $R = 8.0 \, J \, mol^{-1} \, K^{-1}$.	$(Q)$ $7 \, kJ$
$(III)$ One mole of a monatomic ideal gas is compressed adiabatically from volume $V = 1/3 \, m^3$ and pressure $2 \, kPa$ to volume $V/8$.	$(R)$ $4 \, kJ$
$(IV)$ Three moles of a diatomic ideal gas whose molecules can vibrate, is given $9 \, kJ$ of heat and undergoes isobaric expansion.	$(S)$ $5 \, kJ$
	$(T)$ $3 \, kJ$

List-$I$	List-$II$
$(a)$ Isothermal	$(i)$ Pressure constant
$(b)$ Isochoric	$(ii)$ Temperature constant
$(c)$ Adiabatic	$(iii)$ Volume constant
$(d)$ Isobaric	$(iv)$ Heat content is constant

Similar Questions

Match List-$I$ with List-$II$:

List-$I$ List-$II$

$(a)$ Isothermal $(i)$ Pressure constant

$(b)$ Isochoric $(ii)$ Temperature constant

$(c)$ Adiabatic $(iii)$ Volume constant

$(d)$ Isobaric $(iv)$ Heat content is constant

Choose the correct answer from the options given below:

The change in internal energy of a given mass of a gas,when its volume changes from $V$ to $3V$ at constant pressure $P$,is (where $\gamma$ is the ratio of the specific heat capacities of the gas).

For an ideal gas, a cyclic process $ABCA$ as shown in the $P-T$ diagram, when presented in a $P-V$ plot, would be:

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